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    NCERT Exemplar Class 12 Physics Solutions Chapter 9 Ray Optics and Optical Instruments

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    NCERT Exemplar Class 12 Physics Solutions Chapter 9 Ray Optics and Optical Instruments

    Edited By Safeer PP | Updated on Sep 14, 2022 01:03 PM IST | #CBSE Class 12th

    NCERT Exemplar Class 12 Physics solutions chapter 9 is one of the most scoring chapters from the view of exams and gives an insight on various different topics related to visible light. NCERT Exemplar Class 12 Physics chapter 9 solutions deals with various phenomena relating to light and explains the concept of reflection of light by spherical mirrors. Class 12 Physics NCERT Exemplar solutions chapter 9 also contains various mathematical derivations for different terms relating to such concepts, such as for obtaining the focal length of spherical mirrors and mirror equations which help in obtaining different values out of some given information. NCERT chapter ray optics also explains the phenomena of refraction and total internal reflection in detail along with its practical application in the surrounding nature. Students can make use of NCERT Exemplar Class 12 Physics solutions chapter 9 PDF download for preparing for exams.

    Also see - NCERT Solutions for Class 12 Physics

    NCERT Exemplar Class 12 Physics Solutions Chapter 9 MCQI

    Question:1

    A ray of light incident at an angle on a refracting face of a prism emerges from the other face normally. If the angle of the prism is 5° and the prism is made of a material of refractive index 1.5, the angle of incidence is
    A. 7.5^{o}
    B. 5^{o}
    C.15^{o}
    D. 2.5^{o}

    Answer:

    The correct option is (a) 7.5^{o}
    As it is a thin prism, the distance between the refracting surfaces is non-accountable and (A), the prism angle is also very small. Since angle (A) is small, it implies that both r_{1} and r_{2} are also small because of the relation A=r_{1}+r_{2}. This also applies to both i_{1} and i_{2}.

    According to snell's law, 1 \sin i_{1}=\mu \; \sin r_{1}\Rightarrow i_{1}=\mu\; r_{1}
    Also, 1 \sin i_{2}=\mu \; \sin r_{2}\Rightarrow i_{2}=\mu\; r_{2}
    Therefore, deviation, \delta =\left ( i_{1}-r_{1} \right )+\left ( i_{2}-r_{2} \right )
    \Rightarrow \; \; \; \; \delta =\left ( i_{1}+i_{2} \right )-\left ( r_{1}+r_{2} \right )=\left ( r_{1}+r_{2} \right )\left ( \mu -1 \right )
    \Rightarrow \delta =A\left ( \mu -1 \right )
    Since, deviation \delta =A\left ( \mu -1 \right )A
    =\left ( 1.5-1 \right )\times 5^{o}=2.5^{o}
    The angle of the prism is 5^{o}. The ray emr=erges from refracting face of a prism normally.
    Then, i_{2}=r_{2}=0
    As A= r_{1}+r_{2}\Rightarrow r_{i}=A\; or\; r_{1}=5^{o}
    But i_{1}=\mu .r_{1}=\frac{3}{2}\times 5=7.5^{o}

    Question:2

    A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is
    A. blue
    B. green
    C. violet
    D. red

    Answer:

    The emerging colour would be (d) red, because of the relation v=f\lambda , which provides the velocity of a wave. During the changing of the mediums, the frequency of the wave remains constant. So, v\; \alpha \lambda or wavelength is directly proportional to speed.
    As red colour has the highest wavelength in the spectra, it also has the highest speed according to tot eh relation. Hence, when the light travels in the slab, the first colour to emerge is red.

    Question:3

    An object approaches a convergent lens from the left of the lens with a uniform speed 5 m/s and stops at the focus. The image
    A. moves away from the lens with a uniform speed 5 m/s.
    B. moves away from the lens with uniform acceleration.
    C. moves away from the lens with a non-uniform acceleration.
    D. moves towards the lens with a non-uniform acceleration.

    Answer:

    The answer is the option (c).
    The image will moves away from the lens with a non-uniform acceleration), in this case, the objects are moving with a uniform speed of 5 m/s towards a convergent lens from the left, thus with a non-uniform acceleration, the image moves away from the lens. The image starts at a uniform speed but gets accelerated going from 2F to F; the image goes from 2F to infinity. When the image is at 2F, both the object and image will have the same speed.

    Question:4

    A passenger in an aeroplane shall
    A. never see a rainbow.
    B. may see a primary and a secondary rainbow as concentric circles.
    C. may see a primary and a secondary rainbow as concentric arcs.
    D. shall never see a secondary rainbow.

    Answer:

    The answer is the option (b)
    A passenger has an aeroplane has the possibility to (b) may see a primary and a secondary rainbow as concentric circles because the plane is at an extremely high altitude.
    In case of an object moving towards the convex lens with a constant speed from infinity to focus, the image will move slower in the beginning and then faster.
    V_{i}=\left ( \frac{f}{f+u} \right )^{2}V_{o}
    And if an object approaches the lens, the image moves away from the lens with a non-uniform acceleration.

    Question:5

    You are given four sources of light each one providing a light of a single colour – red, blue, green and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is 90°. Which of the following statements is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence?
    A.The beam of red light would undergo total internal reflection.
    B. The beam of red light would bend towards normal while it gets refracted through the second medium.
    C. The beam of blue light would undergo total internal reflection.
    D. The beam of green light would bend away from the normal as it gets refracted through the second medium.

    Answer:

    The answer is the option (c).
    The beam of blue light would undergo total internal reflection, so option (c) is the correct answer.
    According to the Cauchy relationship, smaller the wavelength higher the refractive index and consequently smaller the critical angle

    We know v=f\lambda, the frequency of wave remains unchanged with medium hence v\alpha \lambda
    The critical angle,
    \sin \; C=\frac{1}{\mu }
    Also, velocity of light
    , v\alpha \frac{1}{\mu }
    According to VIBGYOR, among all given sources of light, the blue light have smallest wavelength. As \lambda _{blue}<\lambda _{yellow} hence v _{blue}<v _{yellow} it means \mu _{blue}<\mu _{yellow}
    It means the critical angle for blue is less than yellow colour, the critical angle is least which facilitates total internal reflection for the beam of blue light.

    Question:6

    The radius of curvature of the curved surface of a Plano-convex lens is 20 cm. If the refractive index of the material of the lens be 1.5, it will
    A. act as a convex lens only for the objects that lie on its curved side.
    B. act as a concave lens for the objects that lie on its curved side.
    C. act as a convex lens irrespective of the side on which the object lies.
    D. act as a concave lens irrespective of side on which the object lies.

    Answer:

    The correct answer is option (c) act as a convex lens irrespective of the side on which the object lies
    Key concept : The relation between f,\mu ,R_{1} and R_{2} is known as lens maker's formula and it is
    \frac{1}{f}=\left ( \mu -1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )
    R_{1}=\infty ,R_{2}=-R
    f=\frac{R}{\left ( \mu -1 \right )}
    Here, R=20 cm, \mu =1.5. On substituting the values, we get
    f=\frac{R}{\mu -1}=\frac{20}{15-1}=40\; cm
    As f>0 means converging nature. Therefore, the lens act as a convex lens irrespective of the side on which the object lies.

    Question:7

    The phenomena involved in the reflection of radio waves by ionosphere is similar to
    A. reflection of light by a plane mirror.
    B. total internal reflection of light in air during a mirage.
    C. dispersion of light by water molecules during the formation of a rainbow.
    D. scattering of light by the particles of air.

    Answer:

    The answer is the option (b)
    The phenomena are very similar to total internal reflection of light in the air during a mirage, option (b). The Ionosphere layer of atmosphere around earth is responsible for reflecting the radio waves. And as the angle of incidence is greater than critical, it is very similar to total internal reflection of light in the air during a mirage.

    Question:8

    The direction of ray of light incident on a concave mirror is shown by PQ while directions in which the ray would travel after reflection is shown by four rays marked 1, 2, 3 and 4 (Fig 9.1). Which of the four rays correctly shows the direction of reflected ray?

    Answer:


    (b) The ray PQ incident to the concave mirror passes through the focus F, after reflection, as shown in the diagram is supposed to be parallel to the principal axis.
    Important Note:
    To find out the exact extended graphical location of the image of any object, one can draw any of the given two rays:
    1. Draw a ray reflected through the focus of the mirror, which is initially parallel to the principal axis (1).
    2. When a ray is drawn through the centre of curvature, it reflects back along with itself (3).
    3. Draw a ray that is initiated through the focus; it will be reflected parallel to the principal axis (2).
    4. A ray drawn incident to the pole gets reflected back in a symmetrical manner.

    Question:9

    The optical density of turpentine is higher than that of water while its mass density is lower. Fig 9.2. shows a layer of turpentine floating over water in a container. For which one of the four ray’s incident on turpentine in Fig 9.2, the path shown is correct?

    A. 1
    B. 2
    C. 3
    D. 4

    Answer:


    The correct path followed by the ray of light is an option (b) 2.
    The Snell's law describes the relationship between the angle of incidence and the angle of refraction.
    \mu _{1}\sin \theta_{1}=\mu _{2}\sin \theta_{2}=\text {Constant}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(1)
    Where \mu _{1}\; \text {and}\; \mu _{2} are refractive indices of the two media.

    In this diagram, the ray of light is going from an optically rarer medium to turpentine which an optically denser medium, then the ray bends towards the normal, which means \theta _{1}> \theta _{2}, but when it goes in the opposite direction, i.e. optically denser medium turpentine to rarer medium water, the ray bends away from the normal.

    Question:10

    A car is moving with at a constant speed of 60\; km\; h^{-1} on a straight road. Looking at the rearview mirror, the driver finds that the car following him is at a distance of 100 m and is approaching with a speed of 5\; km\; h^{-1}. In order to keep track of the car in the rear, the driver begins to glance alternatively at the rear and side mirror of his car after every 2 s till the other car overtakes. If the two cars were maintaining their speeds, which of the following statement (s) is/are correct?
    A. The speed of the car in the rear is 65\; km\; h^{-1}.
    B. In the side mirror the car in the rear would appear to approach with a speed of 5\; km\; h^{-1} to the driver of the leading car.
    C. In the rearview mirror the speed of the approaching car would appear to decrease as the distance between the cars decreases.
    D. In the side mirror, the speed of the approaching car would appear to increase as the distance between the cars decreases.

    Answer:

    The answer is the option (d).
    When an object is placed in front of a mirror, all the positions of objects in the front of a mirror, image is virtual, erect and smaller in size. And as the object moves towards the pole, magnification increases and tends to unity at the pole.

    An object moving along the principal axis – on differentiating the mirror formula with respect to time we get,
    \frac{dv}{dt}=\left ( \frac{v}{u} \right )^{2}\frac{du}{dt}=-\left ( \frac{f}{u-f} \right )^{2}\frac{du}{dt}
    Where \frac{dv}{dt} is the velocity of the image moving along the principal axis, and \frac{du}{dt} is the velocity of the object along the principle axis. The negative sign indicates that the image always moves in the direction opposite to that of the object

    Question:11

    There are certain material developed in laboratories which have a negative refractive index (Fig. 9.3). A ray incident from air (medium 1) into such a medium (medium 2) shall follow a path given by
    A.

    Answer:


    The answer is the option (a)
    When a material has a negative refractive index, it follows the Snell's law in an opposite manner. And when an incident ray from one medium (air) falls on them, it is bending just the way shown in option (a) that is on the same side of the normal.

    NCERT Exemplar Class 12 Physics Solutions Chapter 9 MCQII

    Question:12

    Consider an extended object immersed in water contained in a plane trough. When seen from close to the edge of the trough the object looks distorted because
    A. the apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge.
    B. the angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air.
    C. some of the points of the object far away from the edge may not be visible because of total internal reflection.
    D. water in a trough acts as a lens and magnifies the object.

    Answer:

    All three (a, b, c) are the correct options.
    When the ray of light passes from water into the air from the pencil it is refracted, the beam bends away from the normal when going from high to low refractive index.
    In case of light getting refracted from water surface from the submerged object before reaching to the observer, it bends away from the normal and the angle formed by the from the image of the object is smaller than the real angle formed by the object in the air. This also affects the apparent depth as it becomes nearer to the surface of the water when points are close to the edge as compared to the points far from the edge.
    The angle of incident increase while moving right, and it becomes equal to the critical angle. Thus, due to total internal reflection, some point of the object present far from the edge may not be visible.

    Question:13

    A rectangular block of glass ABCD has a refractive index 1.6. A pin is placed midway on the face AB (Fig. 9.4). When observed from the face AD, the pin shall


    A. appear to be near A.
    B. appear to be near D.
    C. appear to be at the centre of AD.
    D. not be seen at all.

    Answer:


    (a, d) option is the right answer because till the angle of incidence is less than the critical angle, i.e. incident angle on AD of the ray from the pin, pin will always seem nearer to the point A.

    But when the angle of incident is larger than the critical angle, the light undergoes total internal reflection and is no visible through AD.

    Question:14

    Between the primary and secondary rainbows, there is a dark band known as Alexandra’s dark band. This is because
    A. light scattered into this region interfere destructively.
    B. there is no light scattered into this region.
    C. light is absorbed in this region.
    D. angle made at the eye by the scattered rays with respect to the incident light of the sun lies between approximately 42° and 50°.

    Answer:

    Formed because of light scattered into the region interfere destructively the dark Alexander's band lie between the primary and secondary rainbow; thus, (a, d) are the correct options. The angle of the primary rainbow in the observer's eyes is from 41° to 42°, and in the secondary rainbow, the angle in the observer's eyes lies between 51° to 54° with respect to the incident light ray.
    Therefore, the overall angle of scattered rays with reference to incident light of the sun ranges from 42° and 50°.

    Question:15

    A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in
    A. a larger angle to be subtended by the object at the eye and hence viewed in greater detail.
    B. the formation of a virtual erect image.
    C. increase in the field of view.
    D. infinite magnification at the near point.

    Answer:

    Right options are - (a, b)
    A magnifying glass simply contains a convex lens of very small focal length.

    For magnification when the final image is formed at D and \infty \left ( \text {i.e}, m_{D}\; \text {and}\; m_{\infty } \right ).
    m_{D}=\left ( 1+\frac{D}{f} \right )_{max} and m_{\infty }=\left (\frac{D}{f} \right )_{min}
    By using a magnifying glass, any objects can be brought closer to the eye as compared to the normal near point. The angle formed by the object in the eye is larger and thus, is more detailed. The image formed is virtual erect and is also enlarged.

    Question:16

    An astronomical refractive telescope has an objective of focal length 20m and an eyepiece of focal length 2cm.
    A. The length of the telescope tube is 20.02m.
    B. The magnification is 1000.
    C. The image formed is inverted.
    D. An objective of a larger aperture will increase the brightness and reduce chromatic aberration of the image.

    Answer:

    The right answers to this question are options (a, b, and c)
    Key concept :

    • Use to see heavenly bodies.
    • f_{objective}>f_{eye\; lens} and d_{objective}>d_{eye\; lens}
    • The intermediate image is real, inverted and small.
    • The final image is virtual, inverted and small.
    • Magnification:
    • m_{D}=-\frac{f_{0}}{f_{e}}\left ( 1+\frac{f_{e}}{D} \right )and m_{\infty }=-\frac{f_{0}}{f_{e}}
    • Length :
    • L_{D}=f_{0}+u_{e}=f_{0}+\frac{f_{e}D}{f_{e}+D}and L_{\infty }=f_{0}+f_{e}
    The length of the telescope tube is
    f_{0}+f_{e}=20+(0.02)=20.02\; m
    Also,
    m=\frac{20}{0.02}=1000
    Also, the image formed is inverted.

    NCERT Exemplar Class 12 Physics Solutions Chapter 9 Very Short Answer

    Question:17

    Will the focal length of a lens for red light be more, same or less than that for blue light?

    Answer:

    The refractive index depends on the colour of light or wavelength of light.
    Cauchy's equation : \mu =A+\frac{B}{\lambda ^{2}}+\frac{C}{\lambda ^{4}}+......
    As \lambda _{red}>\lambda _{blue} hence \mu _{red}<\mu _{blue}
    Hence parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red.
    By lens maker's formula,
    \frac{1}{f}=\left ( \mu -1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )
    The refractive index for red light is less than that for blue light, \mu _{red}<\mu _{blue}
    Hence \frac{1}{f_{red}}<\frac{1}{f_{blue}}\Rightarrow f_{red}>f_{blue}
    Thus, the focal length for red light will be greater than that for blue light.

    Question:18

    The near vision of an average person is 25cm. To view an object with an angular magnification of 10, what should be the power of the microscope?

    Answer:

    In the given question, the least distance of distinct vision of an average person is 25 cm, in order to view an object with magnification 10,
    Here,
    v=D=25\; cm\; and\; u=f
    But the magnification
    m=\frac{v}{u}=\frac{D}{f}
    \Rightarrow f=\frac{D}{m}=\frac{25}{10}=2.5 \; cm=0.025\; m
    But power
    P=\frac{1}{f(in\; m)}=\frac{1}{0.025}=40\; D
    This is the required power of the lens.

    Question:19

    An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the lens is reversed?

    Answer:

    For a given object position if the focal length of the lens does not change, the image position remains unchanged.
    By lens maker's formula,
    \frac{1}{f}=\left ( \mu -1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )
    For this position R_{1} is positive
    and R_{2} is negative. Hence focal length at this position
    \frac{1}{f_{1}}=\left ( \mu -1 \right )\left ( \frac{1}{\left ( +R_{1} \right )}-\frac{1}{\left ( -R_{2} \right )} \right )=\left ( \mu -1 \right )\left ( \frac{1}{R_{1}}+\frac{1}{R_{2}} \right )
    Now the lens is reversed,
    At this position, R_{2} is positive and R_{1} is negative. Hence focal length at this position is
    \frac{1}{f_{2}}=\left ( \mu -1 \right )\left ( \frac{1}{\left ( +R_{2} \right )}-\frac{1}{\left ( -R_{1} \right )} \right )=\left ( \mu -1 \right )\left ( \frac{1}{R_{1}}+\frac{1}{R_{2}} \right )
    We can observe the focal length of the lens does not change in both positions, hence the image position remains unchanged.

    Question:20

    Three immiscible liquids of densities d_{1}>d_{2}>d_{3} and refractive indices \mu _{1}>\mu_{2}>\mu_{3} are put in a beaker. The height of each liquid column is \frac{h}{3}. A dot is made at the bottom of the beaker. For near normal vision, find the apparent depth of the dot.

    Answer:

    Key concept : Coordinate convention: At the first surface (+upward and -ve downward)
    \frac{\mu _{2}}{h'}-\frac{\mu _{1}}{(-h)}=\frac{\mu _{2}-\mu _{1}}{\infty } (infinity because the surface is plane)
    or h'=\frac{\mu _{2}}{\mu _{1}}h. The negative sign shows that it is on the side of the object

    h' is the apparent depth of O after refraction from interface.

    The position of image O after the refraction from surface-1. If seen from u_{2}, the apparent depth is h_{1}

    h_{1}=\frac{\mu _{2}}{\mu _{1}}\frac{h}{3}
    The negative sign indicates that it is on the side of the object.
    Since the image formed by surface-1. It will act as an object for surface-2. If seen from u_{3} , the apparent depth is h_{2} .
    Similarly, the image formed by medium 2, O_{2} acts as an object for medium 3
    h_{2}=-\frac{\mu _{3}}{\mu _{2}}\left (\frac{\mu _{2}}{\mu _{1}}\frac{h}{3}+\frac{h}{3} \right )=-\frac{h}{3}\left (\frac{\mu _{3}}{\mu _{2}}+\frac{\mu _{2}}{\mu _{1}} \right )
    Finally the image formed by surface -2 will act as an object for surface-2. If seen from outside, the apparent depth is h_{3}
    h_{3}=-\frac{1}{\mu _{3}}\left [\frac{h}{3}+ \frac{h}{3}\left (\frac{\mu _{3}}{\mu _{2}}+\frac{\mu _{2}}{\mu _{1}} \right ) \right ]=-\frac{h}{3}\left ( \frac{1}{\mu _{1}}+\frac{1}{\mu _{2}}+\frac{1}{\mu _{3}} \right )
    Hence apparent depth of odt is \frac{h}{3}\left ( \frac{1}{\mu _{1}}+\frac{1}{\mu _{2}}+\frac{1}{\mu _{3}} \right )
    Important point:
    Apparent depth (distance of final image from final surface)
    =\frac{t_{1}}{n_{1\; real}}+\frac{t_{2}}{n_{2\; real}}+\frac{t_{3}}{n_{3\; real}}+...+\frac{t_{n}}{n_{n\; real}}

    Question:21

    For a glass prism \mu=\sqrt{3} the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism.

    Answer:

    The refractive index of prism angle A and angle of minimum deviation \delta _{m} is given by
    \mu=\frac{\sin\left [ \frac{\left ( A+\delta _{m} \right )}{2} \right ]}{\sin\left ( \frac{A}{2} \right )}
    Here we are given, \delta _{m}=A
    Substituting the value, we have
    \mu=\frac{\sin A}{\sin \frac{A}{2}}
    \Rightarrow \mu=\frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{\sin\frac{A}{2}}=2 \cos\frac{A}{2}
    \Rightarrow \mu=\frac{\sin A}{\sin \frac{A}{2}}=\frac{2\sin\frac{A}{2}\cos\frac{A}{2}}{\sin\frac{A}{2}}=2 \cos\frac{A}{2}
    For the given value of refractive index,
    We have, \cos\frac{A}{2}=\frac{\sqrt{3}}{2}\Rightarrow \frac{A}{2}=30^{o}
    Or A=60^{o}
    This is the required value of prism angle.

    NCERT Exemplar Class 12 Physics Solutions Chapter 9 Short Answer

    Question:22

    A short object of length L is placed along the principal axis of a concave mirror away from focus. The object distance is u. If the mirror has a focal length f, what will be the length of the image? You may take L<<\left | v-f \right |

    Answer:

    Thin mirror formula :
    \frac{1}{v}+\frac{1}{u}=\frac{1}{f}
    u= object distance and v= image distance
    Since, the object distance is ,u. Let us consider the two ends of the object be at distance u_{1} and u_{2} respectively, so that du=\left | u_{1}-u_{2} \right |=L. hence size of the image can be written as dv.
    By differentiating both sides
    \left ( -\frac{du}{u^{2}} \right )+\left ( -\frac{dv}{v^{2}} \right )=0\Rightarrow \frac{dv}{v^{2}}=-\frac{du}{u^{2}}
    or
    dv=-\left ( \frac{v^{2}}{u^{2}} \right )du \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(i)
    As \frac{1}{v}=\frac{1}{f}-\frac{1}{u}\Rightarrow v=\frac{fu}{u-f}
    Or \frac{v}{u}=\frac{f}{u-f} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(ii)
    From (i) and (ii
    ) dv=-\left ( \frac{f}{u-f} \right )^{2}du
    But du=L, hence the length of the image is
    \frac{f^{2}}{(u-f)^{2}}L
    This is the required expression for the length of the image.

    Question:23

    A circular disc of radius ‘R’ is placed co-axially and horizontally inside an opaque hemispherical bowl of radius ‘a’ (Fig. 9.5). The far edge of the disc is just visible when viewed from the edge of the bowl. The bowl is filled with transparent liquid of refractive index \mu and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed?

    Answer:


    As shown in the figure, AM is in the direction of the incidence ray before filling the liquid. After filling liquid in the bowl, BM is in the direction of the incident ray. Refracted in both cases is same as that along MN.

    Let the disc is separated by O at a distance d as shown in the figure. Also, considering the angle
    P=90^{o}, OM=a,CB=R,BP=a{-}R,AP=a+R
    Here, in \Delta BMP,
    \sin\; i=\frac{BP}{BM}=\frac{a-R}{\sqrt{d^{2}+\left ( a-R \right )^{2}}}\; \; \; \; \; \; \; \; \; \; \; ....(i)
    and in
    \Delta AMP\; \cos \left ( 90^{o}-\alpha \right )=\sin \; \alpha =\frac{a+R}{\sqrt{d^{2}+\left ( a+R \right )^{2}}}\; \; \; \; \; \; \; \; \; \; \; \; .....(ii)
    But on applying Snell's law at point M
    \mu \times \sin \; i=1 \times \; sin \; r
    \mu \times \frac{a-R}{\sqrt{d^{2}+\left ( a-R \right )^{2}}}=1\times \frac{a+R}{\sqrt{d^{2}+\left ( a+R \right )^{2}}}
    \Rightarrow d=\frac{\mu \left ( a^{2}-b^{2} \right )}{\sqrt{\left ( a+r \right )^{2}-\mu \left ( a-r \right )^{2}}}

    Question:24

    A thin convex lens of focal length 25 cm is cut into two pieces 0.5 cm above the principal axis. The top part is placed at \left ( 0,0 \right ) and an object placed at \left ( -50\; cm,0 \right ). Find the coordinates of the image.

    Answer:

    The focal length will remain the same for each of the parts in case an asymmetric lens is cut in two parts parallel to the principle axis. In such a case the intensity of image that has been formed by every part will be lower in comparison to that of the complete lens.
    If there was no cut, then the object would have been at the height of 0.5 cm from the principal axis OO'

    The top part is placed at \left ( 0,0 \right ) and an object placed at \left ( -50\; cm,0 \right ). There is no effect on the focal length of the lens.

    Applying lens formula,
    \frac{1}{v}-\frac{1}{u}=\frac{1}{f}
    \frac{1}{v}=\frac{1}{u}+\frac{1}{f}=\frac{1}{-50}+\frac{1}{25}=\frac{1}{50}\Rightarrow v=50\; cm
    Magnification is
    m=\frac{v}{u}=-\frac{50}{50}=-1
    Therefore, the image would have been formed ate 50 cm from the pole and 0.5 cm below the principal axis. Thus, with respect to the X-axis passing through the edge of the cut lens, the coordinates of the image are (50\; cm, -1 \; cm).

    Question:25

    In many experimental set-ups the source and screen are fixed at a distance say D and the lens is movable. Show that there are two positions for the lens for which an image is formed on the screen. Find the distance between these points and the ratio of the image sizes for these two points.

    Answer:

    This method is known as the "Displacement method" is also used to find the confocal length of the length in laboratory
    \frac{1}{f}=\frac{1}{v}-\frac{1}{u}
    If we take a convex lens L place between an object O and a screen S. The distance between the object and the screen would be D and the position of the object and screen are held fix. Let the distance of position 1 from the object be x_{1}. Then the distance of the screen from the lens is D-x_{1}.

    Therefore, u=-x_{1} and v=+\left ( D-x_{1} \right ).
    Placing it in the lens formula,
    \frac{1}{D-x_{1}}-\frac{1}{\left ( -x_{1} \right )}=\frac{1}{f}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .....(i)
    At position II, let the distance of the lens from the screen be x_{2}. Then the distance of the screen from the lens is D-x_{2}
    Therefore, u=-x_{2} and v=+\left ( D-x_{2} \right ).
    Placing it in the lens formula
    \frac{1}{D-x_{2}}-\frac{1}{\left ( -x_{2} \right )}=\frac{1}{f}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .....(ii)
    Comparing Eqs. (i) and (ii), we realize that there are only two solutions:
    1.x_{1}=x_{2}; \text {or}
    2.D-x_{1}=x_{2}; \text {and}\; D-x_{2}=x_{1}
    The first solution is trivial. Therefore, if the first position of the lens, for a sharp image, is x_{1} fro the object, the second position is at D-x_{1} from the object.
    Let the distance between the two positions I and II be d.
    From the diagram, it is clear that
    D=x_{1}+x_{2}\; \text {and}\; d=x_{2}-x_{1}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(iii)
    On solving, two equations in (iii) we have
    x_{1}=\frac{D-d}{2} and D-x_{1}=\frac{D+d}{2}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(iv)
    Substituting Eq. (iv) in Eq. (i), we get
    \frac{1}{\left ( \frac{D+d}{2} \right )}-\frac{1}{\left ( -\frac{D-d}{2} \right )}=\frac{1}{f}\Rightarrow f=\frac{D^{2}-d^{2}}{4D}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ...(v)
    or d=\sqrt{D^{2}-4Df}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .....(vi)
    If u=\frac{D}{2}+\frac{d}{2}, then the image is at v=\frac{D}{2}-\frac{d}{2}.
    \therefore The magnification m_{1}=\frac{D-d}{D+d}
    If u=\frac{D-d}{2}, then v=\frac{D+d}{2},
    \therefore The magnification m_{2}=\frac{D+d}{D-d}
    Thus, \frac{m^{2}}{m_{1}}=\left ( \frac{D+d}{D-d} \right )^{2}
    This is the required expression of magnification.

    Question:26

    A jar of height h is filled with a transparent liquid of refractive index \mu (Fig. 9.6). At the center of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when placed on the top surface symmetrically about the center, the dot is invisible.

    Answer:


    Key concept :

    In the above figure, ray 1 strikes the surface at an angle less than critical angle c and gets refracted in the rarer medium. Ray 2 strikes the surface, making an angle greater than the critical angle and gets internally reflected. All the rays strike at the circle of illuminance gets refracted in a rarer medium. The observer in this field will only see the light coming out of from the C.O.I.
    In the above figure, O is the small dot at the bottom of the jar. The ray from the dot emerges out of the circular patch of water surface of diameter AB till the angle of incidence for the rays OA and AON exceeds the critical angle.

    Rays of light incident at an angle greater than i_{C} are totally reflected with in water and consequently cannot emerge out of the water surface.
    As \ sin \; i_{C}=\frac{1}{\mu }\Rightarrow \tan\; i_{C}=\frac{1}{\sqrt{\mu ^{2}-1}}
    Now, \frac{d}{2h}=\tan\; i_{e}
    \Rightarrow d=\frac{2h}{\sqrt{\mu ^{2}-1}}
    This is the required expression of d.

    Question:27

    A myopic adult has a far point at 0.1 m. His power of accommodation is 4 diopters. (i) What power lenses are required to see distant objects? (ii) What is his near point without glasses? (iii) What is his near point with glasses? (Take the image distance from the lens of the eye to the retina to be 2 cm.)

    Answer:

    Key concepts :

    \frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}
    in terms of power P=P_{1}+P_{2}
    (i) If for the normal relaxed eye of an average person, the power at the far point be P_{f}. The required power
    P_{f}=\frac{1}{f}=\frac{1}{0.1}+\frac{1}{0.02}=60\: D
    By the corrective lens, the object distance at the far point is \infty.
    The power required is ,
    P'_{f}=\frac{1}{f'}=\frac{1}{\infty }+\frac{1}{0.02}=50\; D
    Now for eye+lens system, we have the sum of the eye and that of the glasses P_{g}.
    P'_{f}=P_{f}+P_{g}\Rightarrow 50\; D=60D+P_{g}
    Which gives, P_{g}=-10D
    (ii) For the normal eye his power of accommodation is 4D. Let the power of the normal eye for near vision be P_{n}
    Then, 4=P_{n}-P_{f}\; or \; P_{n}=64\; D
    Let his near point be x_{n'} then
    \frac{1}{x_{n}}+\frac{1}{0.02}=64\; or\; \frac{1}{x_{n}}+50=64
    \frac{1}{x_{n}}=14\Rightarrow x_{n}=\frac{1}{14}m=0.07 m
    (iii) With glasses P_{n}^{'}=P{_{f}}^{'}+4=54
    54=\frac{1}{x{_{n}}^{'}}+\frac{1}{0.02}=\frac{1}{x{_{n}}^{'}}+50
    \frac{1}{x{_{n}}^{'}}=4\Rightarrow x{_{n}}^{'}=\frac{1}{4}m=0.25\; m

    NCERT Exemplar Class 12 Physics Solutions Chapter 9 Long Answer

    Question:28

    Show that for a material with refractive index \mu \geq \sqrt{2} light incident at any angle shall be guided along a length perpendicular to the incident face.

    Answer:

    Let the ray incident on face AB at angle i, after refraction, it travels along PQ and then interact with face AC which is perpendicular to the incident face.

    Any ray is guided along AC if the angle ray makes with the face AC \left ( \phi \right ) is greater than the critical angle. From figure
    \phi +r=90^{o}, therefore \sin\; \phi =\cos\; r\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ......(i)
    If \phi is the critical angle it means, \sin\; \phi \geq \frac{1}{\mu }\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ......(ii)
    From (i) and (ii), \cos \; r \geq \frac{1}{\mu ^{2}}\; or\; 1-\cos ^{2}r\leq 1-\frac{1}{\mu ^{2}}
    i.e. , \Rightarrow \sin ^{2}r\leq 1-\frac{1}{\mu ^{2}}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .....(iii)
    Applying Snell's law on face AB,
    1. \sin \; i=\mu \sin\; r
    Or 1. \sin^{2} \; i=\mu^{2} \sin^{2}\; r\Rightarrow \sin^{2}\; r=\frac{1}{\mu ^{2}} \sin^{2}\; i\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; .....(iv)
    From (i) and (ii), \frac{1}{\mu ^{2}}\sin^{2}i\leq 1-\frac{1}{u^{2}}
    Or \sin^{2}i\leq \mu ^{2}-1
    When i=\frac{\pi }{2}, then we have smallest angle \phi.
    If it is greater than the critical angle, then all other angles of incidence shall be more than the critical angle
    Thus, 1\leq \mu ^{2}-1\; or \; \mu ^{2}\geq 2
    \Rightarrow \mu \geq \sqrt{2}. This is the required result.

    Question:29

    The mixture a pure liquid and a solution in a long vertical column (i.e., horizontal dimensions << vertical dimensions) produces diffusion of solute particles and hence a refractive index gradient along the vertical dimension. A ray of light entering the column at right angles to the vertical is deviated from its original path. Find the deviation in travelling a horizontal distance d << h, the height of the column.

    Answer:

    Let us consider a portion of a ray between x and x+dx inside the liquid solution.
    Let us assume that the angle of incidence ray be \theta and
    Let the ray enters the thin column at the height y.
    As a result of the refraction, the ray deviates from its original path and subsequently emerges at x+dx with an angle
    \theta +d\theta and at a height y+dy.

    From Snell's law,
    \mu \left ( y \right )\sin \theta =\mu \left ( y+dy \right )\sin\left ( \theta+d\theta \right )......(i)

    Let refractive index of the liquid at position y be \mu (y)=\mu, then
    \mu\left ( y+dy \right )=\mu+\left ( \frac{d\mu}{dy} \right )dy=\mu+kdy
    Where k=\left ( \frac{d\mu }{dy} \right )= refractive index gradient along the vertical dimension.
    Hence from (i), \mu \sin\theta=\left ( \mu+kdy \right )\sin \left ( \theta+d \theta \right )
    \mu \sin\theta=\left ( \mu+kdy \right ).\left ( \sin \theta.\cos d \theta+\cos \theta. \sin d\theta \right )
    \mu \sin\theta=\left ( \mu+kdy \right ).\left ( \sin \theta.1+\cos \theta.d \theta \right )\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; ....(ii)
    For small angle \sin d \theta \approx d \theta \; and \; \cos d \theta\approx 1
    \mu \sin \theta=\mu \sin \theta+kdy \; \sin \theta+\mu \cos \theta .d\theta+k\; \cos\; \theta dy.d \theta
    kdy \sin \theta +\mu \cos \theta.d\theta=0\Rightarrow d \theta=-\frac{k}{\mu}\tan \theta dy
    But \tan \theta=\frac{dx}{dy}\; and \: k=\left ( \frac{d\mu}{dy} \right )
    d\theta=-\frac{k}{\mu}\left ( \frac{dx}{dy} \right )dy\Rightarrow d\theta=-\frac{k}{\mu}dx
    Integrating both sides, \int_{0}^{\delta }d\theta =-\frac{k}{\mu}dx
    \Rightarrow \delta =-\frac{kd}{\mu }=-\frac{d}{\mu }\left ( \frac{d\mu }{dy} \right )

    Question:32

    (i) Consider a thin lens placed between a source (S) and an observer (O) (Fig. 9.8). Let the thickness of the lens vary as w(b)=w_{o}-\frac{b_{2}}{\alpha } where b is the vertical distance from the pole. w_{o} is a constant. Using Fermat’s principle i.e. the time of transit for a ray between the source and observer is an extremum; find the condition that all paraxial rays starting from the source will converge at a point O on the axis. Find the focal length.
    (ii) A gravitational lens may be assumed to have a varying width of the form

    w(b)=k_{1}ln\left ( \frac{k_{2}}{b} \right ) b_{min}<b<b_{max}
    =k_{1}ln\left ( \frac{k_{2}}{b_{min}} \right )b<b_{min}
    Show that an observer will see an image of a point object as a ring about the center of the lens with an angular radius
    \beta =\sqrt{\frac{\left ( n-1 \right )k_{1}\frac{u}{v}}{u+v}}

    Answer:


    (i) The time taken by ray to travel from S to P_{1} is
    t_{1}=\frac{SP_{1}}{c}=\frac{\sqrt{u^{2}+b^{2}}}{c}
    Or t_{1}=\frac{u}{c}\left ( 1+\frac{1}{2}\frac{b^{2}}{u^{2}} \right ) assuming b<<1
    The time required to travel from P_{1} to O is
    t_{2}=\frac{P_{1}O}{c}=\frac{\sqrt{v^{2}+b^{2}}}{c}=\frac{v}{c}\left ( 1+\frac{1}{2}\frac{b^{2}}{v^{2}} \right )
    The time required to travel through the lens is
    t_{l}=\frac{(n-1)w(b)}{c}
    where n is the refractive index.
    Thus, the total time is
    t=\frac{1}{c}\left [ u+v+\frac{1}{2}b^{2}\left ( \frac{1}{u}+\frac{1}{v} \right )+(n-1)w(b) \right ]
    Put \frac{1}{D}=\frac{1}{u}+\frac{1}{v}
    Then,
    t=\frac{1}{c}\left ( u+v+\frac{1}{2}\frac{b^{2}}{D}+\left ( n-1 \right )\left ( w_{0}+\frac{b^{2}}{\alpha } \right ) \right )
    Fermat's principle gives the time taken should be minimum. For that first derivative should be zero.
    \frac{dt}{db}=0=\frac{b}{CD}-\frac{2(n-1)b}{c\alpha }
    \alpha =2(n-1)D
    Thus, a convergent lens is formed if \alpha =2(n-1)D. This is independent of and hence all paraxial rays from s will converge at O i.e., for rays b<<n\; and\; b<<v
    Since, \frac{1}{D}=\frac{1}{u}+\frac{1}{v}, the focal length is D.
    (ii) In this case, differentiating expression of time taken t e.r.t. b.
    t=\frac{1}{c}\left ( u+v+\frac{1}{2}\frac{b^{2}}{D}+(n-1)k_{1}ln\left ( \frac{k_{2}}{b} \right ) \right )
    \frac{dt}{db}=0=\frac{b}{D}-(n-1)\frac{k_{1}}{b}
    \Rightarrow b^{2}=(n-1)k_{1}D
    \therefore b=\sqrt{(n-1)k_{1}D}
    Thus, all rays passing at a height b shall contribute to the image. The ray paths make an angle.
    \beta =\frac{b}{v}=\frac{\sqrt{(n-1)k_{1}D}}{v}=\sqrt{\frac{(n-1)k_{1}uv}{v^{2}(u+v)}}=\sqrt{\frac{(n-1)k_{1}u}{(u+v)v}} This is the required expression.

    NCERT Exemplar Class 12 Physics Solutions Chapter 9 Ray Optics and Optical Instruments - Main Subtopic

    • Introduction
    • Reflection of light by spherical surfaces
    • Sign convention
    • Focal length of spherical mirrors
    • The Mirror equation
    • Refraction
    • Total internal reflection
    • Total internal reflection in nature and its technological applications
    • Refraction at spherical surface and by lenses
    • Refraction at a spherical surface
    • Refraction by a lens
    • Power of a lens
    • Combination of thin lenses in contact
    • Refraction through a prism
    • Dispersion by a prism
    • Some natural phenomena due to sunlight
    • The rainbow
    • Scattering of light
    • Optical instruments
    • The eye
    • The microscope
    • Telescope

    NCERT Exemplar Class 12 Physics Chapter Wise Links

    Important Topics To Cover For Exams From NCERT Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

    · NCERT Exemplar Solutions For Class 12 Physics chapter 9 covers questions related to the phenomena of refraction by the lenses and how formulas are derived out of it to calculate the Power and combination of lenses required.

    · Class 12 Physics NCERT Exemplar solutions chapter 9 also explores the diagrammatic representation of total internal reflection in different objects of different shapes such as a prism, diamond, etc.

    · Most importantly, NCERT Exemplar Class 12 Physics solutions chapter 9 gives full detailed insight into different optical instruments such as the eye, the microscope, and the telescope, the phenomena associated, the principles used, and the detailed process of such happenings.

    NCERT Exemplar Class 12 Solutions

    Also, check the NCERT solutions of questions given in book

    Also read NCERT Solution subject wise

    Must read NCERT Notes subject wise

    Also Check NCERT Books and NCERT Syllabus here:

    Frequently Asked Question (FAQs)

    1. How important is chapter 9 in physics of Class 12?

    This chapter has the highest weightage in the board exam, thus scoring well in this chapter  will help in scoring in the final paper.

    2. How are these questions solved?

     Each and every question is solved in complete detail covering the basics, the numericals and the diagrams in NCERT exemplar Class 12 Physics solutions chapter 9.

    3. What all things one will learn in Class 12 Physics chapter 9 Ray Optics and Optical Instruments?

    One will cover almost every topic related to ray optics; like reflection, refraction, dispersion, TIR, optical instruments etc.

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    Questions related to CBSE Class 12th

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    hello,

    Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

    I hope this was helpful!

    Good Luck

    Hello dear,

    If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.


    As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.


    Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.


    Believe in Yourself! You can make anything happen


    All the very best.

    Hello Student,

    I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects  and  we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

    You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

    All the best.

    If you'll do hard work then by hard work of 6 months you can achieve your goal but you have to start studying for it dont waste your time its a very important year so please dont waste it otherwise you'll regret.

    Yes, you can take admission in class 12th privately there are many colleges in which you can give 12th privately.

    View All

    A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

    Option 1)

    0.34\; J

    Option 2)

    0.16\; J

    Option 3)

    1.00\; J

    Option 4)

    0.67\; J

    A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

    Option 1)

    2.45×10−3 kg

    Option 2)

     6.45×10−3 kg

    Option 3)

     9.89×10−3 kg

    Option 4)

    12.89×10−3 kg

     

    An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

    Option 1)

    2,000 \; J - 5,000\; J

    Option 2)

    200 \, \, J - 500 \, \, J

    Option 3)

    2\times 10^{5}J-3\times 10^{5}J

    Option 4)

    20,000 \, \, J - 50,000 \, \, J

    A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

    Option 1)

    K/2\,

    Option 2)

    \; K\;

    Option 3)

    zero\;

    Option 4)

    K/4

    In the reaction,

    2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

    Option 1)

    11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

    Option 2)

    6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

    Option 3)

    33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

    Option 4)

    67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

    How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

    Option 1)

    0.02

    Option 2)

    3.125 × 10-2

    Option 3)

    1.25 × 10-2

    Option 4)

    2.5 × 10-2

    If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

    Option 1)

    decrease twice

    Option 2)

    increase two fold

    Option 3)

    remain unchanged

    Option 4)

    be a function of the molecular mass of the substance.

    With increase of temperature, which of these changes?

    Option 1)

    Molality

    Option 2)

    Weight fraction of solute

    Option 3)

    Fraction of solute present in water

    Option 4)

    Mole fraction.

    Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

    Option 1)

    twice that in 60 g carbon

    Option 2)

    6.023 × 1022

    Option 3)

    half that in 8 g He

    Option 4)

    558.5 × 6.023 × 1023

    A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

    Option 1)

    less than 3

    Option 2)

    more than 3 but less than 6

    Option 3)

    more than 6 but less than 9

    Option 4)

    more than 9

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    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    Transportation Planner

    A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

    3 Jobs Available
    Construction Manager

    Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.

    2 Jobs Available
    Carpenter

    Carpenters are typically construction workers. They stay involved in performing many types of construction activities. It includes cutting, fitting and assembling wood.  Carpenters may help in building constructions, bridges, big ships and boats. Here, in the article, we will discuss carpenter career path, carpenter salary, how to become a carpenter, carpenter job outlook.

    2 Jobs Available
    Welder

    An individual who opts for a career as a welder is a professional tradesman who is skilled in creating a fusion between two metal pieces to join it together with the use of a manual or fully automatic welding machine in their welder career path. It is joined by intense heat and gas released between the metal pieces through the welding machine to permanently fix it. 

    2 Jobs Available
    Environmental Engineer

    Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems. 

    2 Jobs Available
    Naval Architect

    A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability. 

    2 Jobs Available
    Welding Engineer

    Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

    2 Jobs Available
    Field Surveyor

    Are you searching for a Field Surveyor Job Description? A Field Surveyor is a professional responsible for conducting field surveys for various places or geographical conditions. He or she collects the required data and information as per the instructions given by senior officials. 

    2 Jobs Available
    Orthotist and Prosthetist

    Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

    6 Jobs Available
    Pathologist

    A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

    5 Jobs Available
    Veterinary Doctor

    A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

    5 Jobs Available
    Gynaecologist

    Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

    4 Jobs Available
    Oncologist

    An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

    3 Jobs Available
    Surgical Technologist

    When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications. 

    Also Read: Career as Nurse

    3 Jobs Available
    Critical Care Specialist

    A career as Critical Care Specialist is responsible for providing the best possible prompt medical care to patients. He or she writes progress notes of patients in records. A Critical Care Specialist also liaises with admitting consultants and proceeds with the follow-up treatments. 

    2 Jobs Available
    Ophthalmologist

    Individuals in the ophthalmologist career in India are trained medically to care for all eye problems and conditions. Some optometric physicians further specialize in a particular area of the eye and are known as sub-specialists who are responsible for taking care of each and every aspect of a patient's eye problem. An ophthalmologist's job description includes performing a variety of tasks such as diagnosing the problem, prescribing medicines, performing eye surgery, recommending eyeglasses, or looking after post-surgery treatment. 

    2 Jobs Available
    Actor

    For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

    4 Jobs Available
    Acrobat

    Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

    3 Jobs Available
    Video Game Designer

    Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

    3 Jobs Available
    Talent Agent

    The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

    If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

    3 Jobs Available
    Radio Jockey

    Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

    A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

    3 Jobs Available
    Social Media Manager

    A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

    2 Jobs Available
    Choreographer

    The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

    2 Jobs Available
    Talent Director

    Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots. 

    2 Jobs Available
    Copy Writer

    In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

    5 Jobs Available
    Editor

    Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

    3 Jobs Available
    Journalist

    Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

    3 Jobs Available
    Publisher

    For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

    3 Jobs Available
    Vlogger

    In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

    3 Jobs Available
    Travel Journalist

    The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

    2 Jobs Available
    Videographer

    Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production. 

    2 Jobs Available
    SEO Analyst

    An SEO Analyst is a web professional who is proficient in the implementation of SEO strategies to target more keywords to improve the reach of the content on search engines. He or she provides support to acquire the goals and success of the client’s campaigns. 

    2 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    Quality Controller

    A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

    A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

    3 Jobs Available
    Production Manager

    Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers. 

    Resource Links for Online MBA 

    3 Jobs Available
    Team Lead

    A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

    2 Jobs Available
    Quality Systems Manager

    A Quality Systems Manager is a professional responsible for developing strategies, processes, policies, standards and systems concerning the company as well as operations of its supply chain. It includes auditing to ensure compliance. It could also be carried out by a third party. 

    2 Jobs Available
    Merchandiser

    A career as a merchandiser requires one to promote specific products and services of one or different brands, to increase the in-house sales of the store. Merchandising job focuses on enticing the customers to enter the store and hence increasing their chances of buying a product. Although the buyer is the one who selects the lines, it all depends on the merchandiser on how much money a buyer will spend, how many lines will be purchased, and what will be the quantity of those lines. In a career as merchandiser, one is required to closely work with the display staff in order to decide in what way a product would be displayed so that sales can be maximised. In small brands or local retail stores, a merchandiser is responsible for both merchandising and buying. 

    2 Jobs Available
    Procurement Manager

    The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness. 

    2 Jobs Available
    Production Planner

    Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner. 

    2 Jobs Available
    ITSM Manager

    ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks. 

    3 Jobs Available
    Information Security Manager

    Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

    3 Jobs Available
    Computer Programmer

    Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

    3 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    IT Consultant

    An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

    2 Jobs Available
    Data Architect

    A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements. 

    2 Jobs Available
    Security Engineer

    The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

    2 Jobs Available
    UX Architect

    A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy. 

     

    2 Jobs Available
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