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NCERT Solutions for Class 12 Physics Chapter 9 - Ray Optics and Optical Instruments
Have you ever thought of how your reflection can be seen in a mirror or of how glasses can make you see? These day-to-day experiences are founded on the principles of Ray optics as explained in Class 12 physics Chapter 9 - Ray optics and optical instruments. Some of the key concepts that are discussed in this chapter include the law of reflection and refraction, total internal reflection, mirror and lens formation, and mechanisms of optical devices such as the human eye, microscope, telescope, and cameras.
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NCERT Solutions for Class 12 Physics Chapter 9 - Ray Optics and Optical Instruments are prepared by experienced subject experts according to the latest CBSE syllabus. These NCERT Solutions for Class 12 Physics Chapter 9 - Ray Optics and Optical Instruments answer all NCERT exercise questions in a proper step-by-step manner, and assist the students to gain solid conceptual clarity and enhance their numerical-solving capabilities. These thoroughly constructed NCERT solutions are very practical in the preparation for CBSE Class 12 board exams and competitive examinations such as JEE and NEET, where knowledge of optical concepts and derivations is a must.
Ray Optics and Optical Instruments NCERT Solutions: Download PDF
The Class 12 Physics Chapter 9 - Ray Optics and Optical Instruments question answers explain reflection, refraction, and the work or mechanism of lenses and mirrors in a simple way. These step-by-step Class 12 physics chapter 9 Ray Optics and Optical Instruments question answers are very beneficial in board exams and competitive exams such as JEE and NEET. Students may also download the free PDF of NCERT Solutions for Class 12 Physics Chapter 9 to revise and rehearse it anywhere, improving their preparation.
Ray Optics and Optical Instruments NCERT Solutions: Exercise Questions
Class 12 Physics Chapter 9 - Ray Optics and Optical Instruments exercise questions assist students in training fundamental concepts such as image formation, lens, and mirror equations and the functioning of optical instruments. The Ray Optics and Optical Instruments class 12 question answers offer clear and step-wise solutions that enhance conceptual clarity and exam preparation.
Answer:
Given the size of the candle, h = 2.5 cm
Object distance, u = 27 cm
The radius of curvature of the concave mirror, R = -36 cm
focal length of a concave mirror = R/2 = -18 cm
let image distance = v
Now, as we know
$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
$\frac{1}{-27}+\frac{1}{v}=\frac{1}{-18}$
$\frac{1}{v}=\frac{1}{-18} + \frac{1}{27}$
$v= -54cm$
Now, let the height of the image be $h'$
The magnification of the image is given by
$m=\frac{h'}{h}=-\frac{v}{u}$
from here
${h'}=-\frac{-54}{-27}*2.5=-5cm$
Hence, the size of the image will be -5cm. The negative sign implies that the image is inverted and real.
If the candle is moved closer to the mirror, we have to move the screen away from the mirror in order to obtain the image on the screen. If the image distance is less than the focal length image cannot be obtained on the screen, and the image will be virtual.
Answer:
Given the height of the needle, h = 4.5 cm
distance of object = 12 cm
focal length of convex mirror = 15 cm.
Let the distance of the image be v
Now, as we know
$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f} - \frac{1}{u}$
$\frac{1}{v}=\frac{1}{15} - \frac{1}{-12}$
$\frac{1}{v}=\frac{1}{15} + \frac{1}{12}$
v = 6.7 cm
Hence, the distance of the image is 6.7 cm from the mirror, and it is on the other side of the mirror.
Now, let the size of the image be h'
so.
$m=-\frac{v}{u} = \frac{h'}{h}$
$h'= -\frac{v}{u}*h$
$h'= -\frac{6.7}{-12}*4.5$
$h'= 2.5 cm$
Hence, the size of the image is 2.5 cm. The positive sign implies the image is erect, virtual and diminished.
Magnification of the image = $\frac{h'}{h}$ = $\frac{2.5}{4.5}$ = 0.56
m = 0.56
The image will also move away from the mirror if we move the needle away from the mirror, and the size of the image will decrease gradually.
Answer:
Given:
Actual height of the tank,h = 12.5 cm
Apparent height of tank,h' = 9.4 cm
let refrective index of the water be $\mu$
$\mu = \frac{h}{h'} =\frac{12.5}{9.4} = 1.33 (approx)$
So the refractive index of water is approximately 1.33.
Now, when water is replaced with a liquid having $\mu = 1.63$
$\mu = \frac{h}{h'} =\frac{12.5}{h'_{new}} = 1.63$
$h'_{new}= \frac{12.5}{1.63}= 7.67 cm$
Hence, the new apparent height of the needle is 7.67 cm.
Total distance we have to move in a microscope = 9.4 - 7.67 = 1.73 cm.
Since the new apparent height is less than the previous apparent height, we have to move UP the microscope in order to focus the needle.
Answer:
As we know, by Snell's law
$\mu_1\sin\theta _1=\mu _2\sin\theta _2$ where,
$\mu_1$ = refractive index of medium 1
$\theta _1$ = incident angle in medium 1
$\mu _2$ = refrective index of medium 2
$\theta _2$ = refraction angle in medium 2
Now, applying it to fig (a)
$1\sin 60=\mu _{glass}\sin 35$
$\mu _{glass}=\frac{\sin 60}{\sin 35}=\frac{0.866025}{0.573576 } = 1.509$
Now applying for fig (b)
$1\sin 60=\mu _{water}\sin 47$
$\mu _{water}=\frac{\sin 60}{\sin 47}=\frac{.8660}{.7313} = 1.184$
Now in fig (c) let the refraction angle be $\theta$ so,
$\mu _{water}\sin 45=\mu _{glass}\sin\theta$
$\sin\theta =\frac{\mu _{water}*\sin 45}{\mu _{glass}}$
$\sin\theta =\frac{1.184*0.707}{1.509} = 0.5546$
$\theta = \sin^{-1}(0.5546) = 38.68$
Therefore, the angle of refraction when the ray goes from water to glass in Figure (c) is 38.68.
Answer:
Rays of light will emerge in all directions and up to the angle when total internal reflection starts, i.e. when the angle of refraction is 90 degrees.
Let the incident angle be i when the refraction angle is 90 degrees.
So, by Snell's law
$\mu _{water}\sin i=1\sin 90$
from here, we get
$\sin i=\frac{1}{1.33}$
$i=\sin^{-1}(\frac{1}{1.33})=48.75^0$
Now, let R be the Radius of the circle of the area from which the rays are emerging. and d be the depth of water, which is = 80 cm.
From the figure:
$tani=\frac{R}{d}$
$R = tani*d=tan48.75^0*80cm$
R = 91 cm
So the area of the water surface through which the rays will be emerging is
$\Pi R^2 = 3.14*(91)^2cm^2$
$= 2.61m^2$
therefore required area $= 2.61m^2$ .
Answer:
In Prism :
Prism angle ( $A$ ) = First Refraction Angle ( $r _1$ ) + Second refraction angle ( $r _2$ )
also, Deviation angle ( $\delta$ ) = incident angle( $i$ ) + emerging angle( $e$ ) - Prism angle ( $A$ ) ..............(1)
the deviation angle is minimum when the incident angle( $i$ ) and an emerging angle( $e$ ) are the same. In other words
$i=e$ ...........(2)
from (1) and (2)
$\delta_{min} = 2i-A$
$i=\frac{\delta_{min} +A}{2}$ ..........................(3)
We also have
$r _1 = r_2 = r = \frac{A}{2}$ .................(4)
Now applying Snell's law using equations (3) and (4)
$\mu _1\sin i=\mu _2\sin r$
$1\sin(\frac{\delta _{min}+A}{2})=\mu _2\sin\frac{A}{2}$
$\mu _2=\frac{\sin(\frac{\delta _{min}+A}{2})}{\sin\frac{A}{2}}$ ...................(5)
Given
$\\\delta _{min}= 40 \\A = 60$
Putting those values in (5), we get
$\mu _2=\frac{\sin(\frac{40+60}{2})}{\sin\frac{60}{2}}=\frac{\sin 50}{\sin 30}=1.532$
Hence, the refractive index of the prism is 1.532.
Now, when the prism is in the water.
Applying Snell's law:
$\mu _1\sin(\frac{\delta _{min}+A}{2})=\mu _2\sin\frac{A}{2}$
$1.33\sin(\frac{\delta _{min}+60}{2})=1.532\sin\frac{60}{2}$
$\sin(\frac{\delta _{min}+60}{2})=\frac{1.532*0.5}{1.33}$
$\frac{\delta _{min}+60}{2}=\sin^{-1}\frac{1.532*0.5}{1.33}$
$\delta _{min} =2\sin^{-1}0.5759 - 60$
$\delta _{min} =2*35.16- 60 =10.32^0$
Hence minimum angle of deviation inside water is 10.32 degrees.
Answer:
As we know, the lens maker's formula
$\frac{1}{f}=(\mu _{21}-1)(\frac{1}{R_1}-\frac{1}{R_2})$
[ This is derived by considering the case when the object is at infinity and the image is at the focus]
Where $f$ = focal length of the lens
$\mu _{21}$ = refractive index of the glass of lens with the medium(here air)
$R_1$ and $R_2$ are the Radius of curvature of the faces of the lens.
Here,
Given, $f$ = 20cm,
$R_1$ = $R$ and $R_2$ = $-R$
$\mu _{21}$ = 1.55
Putting these values in the equation,
$\frac{1}{20}=(1.55-1)(\frac{1}{R}-\frac{1}{-R})$
$\frac{2}{R}=\frac{1}{20}*\frac{1}{0.55}$
$R = 40*0.55$
$R = 22cm$
Hence Radius of curvature of the lens will be 22 cm.
Answer:
In any Lens :
$\frac{1}{v} - \frac{1}{u}=\frac{1}{f}$
$\\v=$ the distance of the image from the optical centre
$\\u=$ the distance of the object from the optical centre
$\\f=$ the focal length of the lens
a)
Here, the beam converges from the convex lens to point P. This image P will now act as an object for the new lens, which is placed 12 cm from it, with a focal length of 20 cm.
So,
$\frac{1}{v} - \frac{1}{12}=\frac{1}{20}$
$\frac{1}{v}=\frac{1}{20} + \frac{1}{12}$
$\frac{1}{v}=\frac{8}{60}$
$v = 7.5 cm$
Hence distance of the image is 7.5 cm, and it will form towards the right as the positive sign suggests.
b)
Here, Focal length $f$ = -16cm
so,
$\frac{1}{v} - \frac{1}{12}=\frac{1}{-16}$
$\frac{1}{v} =\frac{1}{-16} + \frac{1}{12} = \frac{1}{48}$
$v = 48 cm$
Hence image distance will be 48 cm in this case, and it will be in the right direction(as the positive sign suggests)
Answer:
In any Lens;
$\frac{1}{v} - \frac{1}{u}=\frac{1}{f}$
$\\v=$ the distance of the image from the optical centre
$\\u=$ the distance of the object from the optical centre
$\\f=$ the focal length of the lens
Here given,
$\\u=$ -14 cm
$\\f=$ -21 cm
$\frac{1}{v} - \frac{1}{-14}=\frac{1}{-21}$
$\frac{1}{v} =\frac{1}{-21}- \frac{1}{14} = \frac{-5}{42}$
$v = -\frac{42}{5}= -8.4cm$
Hence image distance is -8.4 cm. The negative sign indicates the image is erect and virtual.
Also, as we know,
$m= -\frac{v}{u} = \frac{h'}{h}$
From Here
$h'= -\frac{v}{u}*h$
$h'= -\frac{-8.4}{-12}*3= 1.8 cm$
Hence, the height of the image is 1.8 cm.
As we move the object further away from the lens, the image will shift toward the focus of the lens, but will never go beyond that. The size of the object will decrease as we move away from the lens.
Answer:
When two lenses are in contact, the equivalent is given by
$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$
where $f_1$ and $f_2$ are the focal lengths of two individual lenses.
So, Given,
$f_1 =$ 30 cm and $f_2 = -20cm$ (as the focal length of the convex lens is positive and of the concave lens is negative by convention)
Putting these values we get,
$\frac{1}{f}=\frac{1}{30}+\frac{1}{-20}$
$\frac{1}{f}=-\frac{1}{60}$
$f = -60cm$
Hence equivalent focal length will be -60 cm, and since it is negative, the equivalent is behaving as a concave lens, which is also called a diverging lens.
Answer:
In a compound microscope, first, the image of an object is made by the objective lens, and then this image acts as an object for the eyepiece lens.
Given
the focal length of objective lens = $f_{objective}$ = 2 cm
focal length of eyepiece lense = $f_{eyepiece}$ = 6.25cm
Distance between the objective lens and eyepiece lens = 15 cm
a)
Now in the Eyepiece lens
Image distance = $v_{final}$ = -25 cm (least distance of vision with sign convention)
focal length = $f_{eyepiece}$ = 6.25 cm
$\frac{1}{f_{eyepiece}} = \frac{1}{v_{final}} - \frac{1}{u}$
$\frac{1}{u} = \frac{1}{v_{final}} -\frac{1}{f_{eyepiece}}$
$\frac{1}{u} = \frac{1}{-25} -\frac{1}{6.25} = -\frac{1}{5}$
$u = -5 cm.$
Now, this object distance $u$ is from the eyepiece lens; since the distance between lenses is given, we can calculate this distance from the objective lens.
the distance of $u$ from objective lens = $d+u=$ $15 - 5=10cm$ . This length will serve as the image distance for the objective lens.
$v = 10 cm$
So in the objective lens
$\frac{1}{f_{objective}} = \frac{1}{v} - \frac{1}{u_{initial}}$
$\frac{1}{u_{initial}} = \frac{1}{v} - \frac{1}{f_{objective}}$
$\frac{1}{u_{initial}} = \frac{1}{10} - \frac{1}{2} = -\frac{4}{10}=-\frac{2}{5}$
$u_{intial}$ = -2.5 cm
Hence, the object distance required is -2.5 cm.
Now, the magnifying power of a microscope is given by
$m=\frac{v}{|u_{initial}|}(1+\frac{d}{f_{eyepiece}})$ where $d$ is the least distance of vision
So putting these values
$m=\frac{10}{2.5}(1+\frac{25}{6.25}) = 20$
Hence, the lens can magnify the object 20 times.
b) When the image is formed at infinity
in the eyepiece lens,
$\frac{1}{f_{eyepiece}} = \frac{1}{v_{final}} - \frac{1}{u}$
$\frac{1}{6.25} = \frac{1}{infinity} - \frac{1}{u}$
from here $u =$ - 6.25., this distance from objective lens = $d+u$ = 15 - 6.25 = 8.75 = $v$
in the optical lens:
$\frac{1}{f_{objective}} = \frac{1}{v} - \frac{1}{u_{initial}}$
$\frac{1}{2} = \frac{1}{-6.25} - \frac{1}{u_{initial}}$
$\frac{1}{u_{initial}} =- \frac{6.75}{17.5}$
$u_{initial}=-2.59cm$
Now,
$m=\frac{v}{|u_{initial}|}(1+\frac{d}{f_{eyepiece}})$ where $d$ is the least distance of vision
Putting the values, we get,
$m=\frac{8.75}{2.59}(1+\frac{25}{6.25})= 13.51$
Hence magnifying power, in this case, is 13.51.
Q 9.12 A person with a normal near point (25 cm) using a compound microscope with an objective of focal length 8.0 mm and an eyepiece of focal length 2.5cm can bring an object placed at 9.0mm from the objective into sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope
Answer:
Inside a microscope,
For the eyepiece lens,
$\frac{1}{f_{eyepiece}}=\frac{1}{v_{eyepiece}} - \frac{1}{u_{eyepiece}}$
We are given
$v_{eyepiece}= -25cm$
$f_{eyepiece}= 2.5cm$
$\frac{1}{2.5}=\frac{1}{-25} - \frac{1}{u_{eyepiece}}$
$\frac{1}{u_{eyepiece}}=\frac{1}{-25} - \frac{1}{2.5}=-\frac{11}{25}$
$u_{eyepiece}=- \frac{25}{11}=-2.27cm$
We can also find this value by finding the image distance in the objective lens.
So, in the objective lens
$\frac{1}{f_{objective}}=\frac{1}{v_{objective}} - \frac{1}{u_{objective}}$
We are given
$f_{objective}= 0.8$
$u_{objective}= -0.9$
$\frac{1}{0.8}=\frac{1}{v_{objective}} - \frac{1}{-0.9}$
$\frac{1}{v_{objective}} = \frac{0.1}{0.72}$
$v_{objective} = 7.2cm$
Distance between object lens and eyepiece = $|u_{eyepiece}| + v_{objective}$ = 2.27 + 7.2 = 9.47 cm.
Now,
Magnifying power :
$m = \frac{v_{obejective}}{|u_{objective}|}(1+\frac{d}{f_{eyepiece}})$
$m = \frac{7.2}{0.9}(1+\frac{25}{2.5})= 88$
Hence magnifying power for this case will be 88.
Answer:
The magnifying power of the telescope is given by
$m=\frac{f_{objective}}{f_{eyepiece}}$
Here, given,
focal length of objective lens = $f_{objective}=$ 144 cm
focal length of eyepiece lens = $f_{eyepiece}=$ 6 cm
$m=\frac{f_{objective}}{f_{eyepiece}}=\frac{144}{6}=24$
Hence magnifying power of the telescope is 24.
In the telescope distance between the objective and eyepiece is given by
$d= f_{objective}+ f_{eyepiece}$
$d = 144+6=150$
Therefore, the distance between the two lenses is 250 cm.
Answer:
Angular magnification in the telescope is given by :
angular magnification = $\alpha =$ $\frac{f_{objective} }{f_{eyepiece}}$
Here given,
focal length of objective length = 15m = 1500cm
The focal length of the eyepiece = 1 cm
so, angular magnification, $\alpha =$ $\frac{1500}{1}$
$\alpha = 1500$
Answer:
Given,
The radius of the lunar orbit,r = $3.8 \times 10^{8}m$.
The diameter of the moon,d = $3.48 \times 10^{6}m$
focal length $f = 15m$
Let $d_1$ be the diameter of the image of the moon which is formed by the objective lens.
Now,
The angle subtended by the diameter of the moon will be equal to the angle subtended by the image,
$\frac{d}{r}=\frac{d_1}{f}$
$\frac{3.48*10^6}{3.8*10^8}=\frac{d_1}{15}$
$d_1=13.74 cm$
Hence, the required diameter is 13.74cm.
Q 9.15 (a) Use the mirror equation to deduce that:
An object placed between f and 2f of a concave mirror produces a real image beyond 2f.
Answer:
The equation we have for a mirror is:
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$\frac{1}{u}=\frac{1}{f}-\frac{1}{v}$
Given condition $f<u<2f$ and $v>2f$
$\frac{1}{2f}<\frac{1}{u}<\frac{1}{f}$ and $\frac{1}{v}<\frac{1}{2f}$
$-\frac{1}{2f}>-\frac{1}{u}>-\frac{1}{f}$
$\frac{1}{f}-\frac{1}{2f}>\frac{1}{f}-\frac{1}{u}>\frac{1}{f}-\frac{1}{f}$
$\frac{1}{2f}>\frac{1}{v}>0$
${2f}<{v}<0$
Here, $f$ has to be negative in order to satisfy the equation, and hence we conclude that our mirror is a concave Mirror. It also satisfies that $-v>-2f$ (image lies beyond 2f)
Q. 9.15 (b) Use the mirror equation to deduce that:
A convex mirror always produces a virtual image independent of the location of the object.
Answer:
In a convex mirror focal length is positive conventionally.
So we have the mirror equation
$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$
$\frac{1}{v}= \frac{1}{f}-\frac{1}{u}$
Here, since object distance is always negative whenever we put our object on the left side of the convex mirror(which we always do, generally). So $\frac{1}{v}$ is always the sum of two positive quantity(negative sign in the equation and negative sign of the $u$ will always make positive) and hence we conclude that $v$ is always greater than zero which means the image is always on the right side of the mirror which means it is a virtual image. Therefore, a convex lens will always produce a virtual image regardless of anything.
Q 9.15 (c) Use the mirror equation to deduce that:
Answer:
In a convex mirror focal length is positive conventionally.
So we have the mirror equation
$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$
$\frac{1}{v}= \frac{1}{f}-\frac{1}{u}$
here since $f$ is positive and $u$ is negative (conventionally), so we have,
$\frac{1}{v}>\frac{1}{f}$ that is '
$v<f$
which means the image will always lie between the pole and the focus.
Now,
$\frac{1}{v}= \frac{1}{f}-\frac{1}{u}=\frac{u-f}{uf}$
$magnification (m)=-\frac{v}{u} = \frac{f}{f-u}$
Here, since $u$ is always negative conventionally, it can be seen that magnification of the image will always be less than 1, and hence we conclude that the image will always be diminished.
Q 9.15 (d) Use the mirror equation to deduce that:
Answer:
The focal length $f$ of a concave mirror is always negative.
Also conventionally object distance $u$ is always negative.
So we have the mirror equation:
$\frac{1}{f} = \frac{1}{v}+\frac{1}{u}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
Now, in this equation, whenever $u<f$, $\frac{1}{v}$ will always be positive, which means $v$ is always positive, which means it lies on the right side of the mirror, which means the image is always virtual.
Now,
$m=-\frac{v}{u}=-\frac{f}{u-f}$
Since the denominator is always less than the numerator, the magnitude magnification will always be greater than 1
Hence, we conclude that the image is always gonna be enlarged.
Hence, an object placed between the pole and the focus of a concave mirror produces a virtual and enlarged image.
Answer:
As we know,
Refractive index = $\frac{actualdepth}{apparentdepth}$
Here actual depth = 15cm
let apparent depth be d'
And the refractive index of the glass = 1.5
Now putting these values, we get,
$1.5 = \frac{15}{d'}$
$d' =10$
the change in the apparent depth = 15 - 10 = 5 cm.
As long as we are not taking the slab away from the line of sight of the pin, the apparent depth does not depend on the location of the slab.
Answer:
We are given,
Refractive index of glass( $\mu _{glass}$ ) and outer covering( $\mu _{outerlayer}$ ) is 1.68 and 1.44 respectively.
Now applying Snell's law on the upper glass - outer layer,
$\mu _{glass}\sin i'=\mu_{outerlayer}\sin 90$
$i' =$ the angle from where total Internal reflection starts
$\sin i'=\frac{\mu_{outerlayer}}{\mu_{glass}} = \frac{1.44}{1.68}=0.8571$
$i' = 59^0$
At this angle, in the air-glass interface
Refraction angle $r$ = 90 - 59 = 31 degree
Let Incident Angle be $i$.
Applying Snell's law
$1\sin i=\mu_{glass}\sin r$
$\sin i=1.68\sin 31=0.8652$
$i=60$ (approx)
Hence total range of incident angles for which total internal reflection happens is $0<i<60$
Q 9.17 (b) What is the answer if there is no outer covering of the pipe?
Answer:
In the case when there is no outer layer,
Snell's law at the glass-air interface(when the ray is emerging from the pipe)
$\mu _{glass}\sin i'=1\sin 90$
$\sin i'=\frac{1}{\mu_{glass}} = \frac{1}{1.68}=0.595$
$i'=$ 36.5
refractive angle $r$ corresponding to this = 90 - 36.5 = 53.5.
The angle r is greater than the critical angle
So for all of the incident angles, the rays will get totally internally reflected. In other words, rays won't bend in air-glass interference; they would rather hit the glass-air interference and get reflected
Answer:
As we know, for a real image, the maximum focal length is given by
$f_{max}= \frac{d}{4}$
where d is the distance between the object and the lens.
So putting values we get,
$f_{max}= \frac{3}{4}=0.75$
Hence maximum focal length required is 0.75.
Answer:
As we know, the relation between focal length $f$, the distance between the screen $D$ and the distance between the two locations of the object $d$ is :
$f=\frac{D^2-d^2}{4D}$
Given: $D$ = 90 cm., $d$ = 20 cm ,
so
$f=\frac{90^2-20^2}{4*90}$
$f=\frac{90^2-20^2}{4*90}=\frac{770}{36}=21.39cm$
Hence, the focal length of the convex lens is 21.39 cm.
Q 9.20 (a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of the effective focal length of this system useful at all?
Answer:
Here, there are two cases, the first one is the one when we see it from the convex side, i.e. Light is coming from infinity and going into the convex lens first and then goes to the concave lens afterwards. The second case is just the reverse of the first case, i.e. light rays are going into the concave mirror first.
1)When light is incident on a convex lens first
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$
$\frac{1}{v}=\frac{1}{30}+\frac{1}{infinite}$
$v= 30cm$
Now this will act as an object for the concave lens.
$\frac{1}{f_{concave}}=\frac{1}{v_{fromconcave}}-\frac{1}{u_{fromconcave}}$
$u_{fromconcave}= 30 - 8 = 22 cm$
$\frac{1}{-20}=\frac{1}{v_{fromconcave}}-\frac{1}{22}$
$\frac{1}{v}=-\frac{1}{220}$
$v = -220cm$
Hence parallel beam of rays will diverge from this point, which is (220 - 4 = 216) cm away from the centre of the two lenses.
2) When rays fall on the concave lens first
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$
$\frac{1}{v}=\frac{1}{-20}+\frac{1}{infinite}$
$v=-20cm$
Now this will act as an object for a convex lens.
$\frac{1}{f_{convex}}=\frac{1}{v_{fromconvex}}-\frac{1}{u_{fromconvex}}$
$u_{fromconvex}= -20-8=-28cm$
$\frac{1}{-30}=\frac{1}{v_{fromconvex}}-\frac{1}{-28}$
$\frac{1}{v_{fromconvex}}=\frac{1}{30}-\frac{1}{28}=-\frac{1}{420}$
$v_{fromconvex}= -420cm$
Hence parallel beam will diverge from this point, which is (420 - 4 = 146 cm ) away from the centre of the two lenses.
As we have seen for both cases, we have different answers, so yes, answers depend on the side of incidence when we talk about combining lenses. i.e. we can not use the effective focal length concept here.
Answer:
Given
Object height = 1.5 cm
Object distance from convex lens = - 40cm
According to the lens formula
$\frac{1}{v_{fromconvex}}=\frac{1}{f_{convex}}+\frac{1}{u_{fromconvex}}$
$\frac{1}{v_{fromconvex}}=\frac{1}{30}+\frac{1}{-40}=\frac{1}{120}$
$v_{fromconvex}=120$
Magnification due to a convex lens:
$m_{convex}=-\frac{v}{u}=-\frac{120}{-40}=3$
The image of the convex lens will act as an object for the concave lens,
so,
$\frac{1}{v_{fromconcave}}=\frac{1}{f_{concave}}+\frac{1}{u_{fromconcave}}$
$u_{concave}= 120 - 8 = 112$
$\frac{1}{v_{fromconcave}}= \frac{1}{-20}+\frac{1}{112}$
$\frac{1}{v_{fromconcave}}= -\frac{92}{2240}$
$v_{fromconcave}= \frac{-2240}{92}$
Magnification due to a concave lens :
$m_{concave}= \frac{2240}{92}*\frac{1}{112}=\frac{20}{92}$
The combined magnification:
$m_{combined}=m_{convex}*m_{concave}$
$m_{combined}=3*\frac{20}{92}=0.652$
Hence height of the image = $m_{combined}*h$
= 0.652 * 1.5 = 0.98cm
Hence height of the image is 0.98cm.
Answer:
Let the prism be ABC,
as emergent angle $e=90^0$,
$\mu_{glass}\sin r_2 = 1\sin 90$
$\sin r_2=\frac{1}{1.524} = 0.6562$
$r_2= 41 ^0$ (approx)
Now, as we know in the prism
$r_1+r_2=A$
Hence, $r_1=A-r_2=60-41=19^0$
Now applying Snell's law at surface AB
$1\sin i=\mu _{glass}\sin r_1$
$\sin i=1.524\sin 19$
$sini=0.496$
$i = 29.75^0$
Hence, the angle of incidence is 29.75 degrees.
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
Answer:
Given,
Object distance u = -9cm
Focal length of convex lens = 10cm
According to the lens formula
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{10}=\frac{1}{v}-\frac{1}{-9}$
$\\\frac{1}{v}=\frac{1}{10}-\frac{1}{9}\\\Rightarrow v=-90\ cm$
a) Magnification
$m=\frac{v}{u}=\frac{-90}{9}=10 \ cm$
The area of each square in the virtual image
$=10\times10\times1=100mm^2=1cm^2$
b) Magnifying power
$=\frac{d}{|u|}=\frac{25}{9}=2.8$
c) No,
$magnification = \frac{v}{u}$
$magnifying\ power = \frac{d}{|u|}$ .
Both the quantities will be equal only when the image is located at the near point |v| = 25 cm
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case?
Answer:
a)maximum magnification is possible when our image distance is equal to the minimum vision point, that is,
$v=-25$
$f = 10cm$ (Given)
Now, according to the lens formula
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{u}=\frac{1}{v}-\frac{1}{f}$
$\frac{1}{u}=\frac{1}{-25}-\frac{1}{10}$
$\frac{1}{u}=-\frac{7}{50}$
$u=-\frac{50}{7}=-7.14cm$
Hence required object distance for viewing squares distinctly is 7.14 cm away from the lens.
b)
Magnification of the lens:
$m = \left | \frac{v}{u} \right |=\frac{25}{50}*7= 3.5$
c)
Magnifying power
$M = \frac{d}{u} =\frac{25}{50}*7= 3.5$
Since the image is forming at the near point ( d = 25 cm ), both magnifying power and magnification are the same.
Answer:
Given
Virtual image area = 6.25 $mm^{2}$
Actual ara = 1 $mm^{2}$
We can calculate linear magnification as
$m=\sqrt{\frac{6.25}{1}}=2.5$
We also know
$m=\frac{v}{u}$
$v = mu$
Now, according to the lens formula
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{10}=\frac{1}{mu}-\frac{1}{u}$
$\frac{1}{u}(\frac{1}{2.5}-1)=\frac{1}{10}$
$u=-6cm$ and
$v=mu=2.5*(-6)=-15cm$
Since the image is forming at a distance of less than 25 cm, it can not be seen by eye distinctly.
Answer:
Angular magnification is the ratio of tangents of the angle formed by the object and image from the centre point of the lens. In this question angle formed by the object and a virtual image is the same, but it provides magnification in a way that, whenever we have an object placed before 25cm, the lens magnifies it and makes it in the vision range. By using magnification, we can put the object closer to the eye and still see it, which we couldn't have done without magnification.
Answer the question
Answer:
Yes, angular magnification will change if we move our eye away from the lens. This is because the angle subtended by the lens would be different from the angle subtended by the eye. When we move our eye from the lens, angular magnification decreases. Also, one more important point here is that object distance does not have any effect on angular magnification.
Answer:
Firstly, grinding a lens with a very small focal length is not easy, and secondly and more importantly, when we reduce the focal length of a lens, spherical and chromatic aberrations become more noticeable. They are defects of the image, resulting from the way the rays of light.
Q 9.25 (d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
Answer:
We need more magnifying power and angular magnifying power in a microscope in order to use it effectively. Keeping both the objective focal length and eyepiece focal length small makes the magnifying power greater and more effective.
Answer:
When we view through a compound microscope, our eyes should be positioned a short distance away from the eyepiece lens for seeing a clearer image. The image of the objective lens in the eyepiece lens is the position for best viewing. It is also called "eye-ring", and all reflected rays from the lens pass through it, which makes it the ideal position for the eye for the best view.
When we put our eyes too close to the eyepiece lens, then we catch the lesser refracted rays from the eyes, i.e. we reduce our field of view, which affects the clarity of the image.
Answer:
Given,
magnifying power = 30
objective lens focal length
$f_{objeective}$ = 1.25cm
eyepiece lens focal length
$f_{eyepiece}$ = 5 cm
Normally, the image is formed at a distance d = 25cm
Now, by the formula;
Angular magnification by eyepiece:
$m_{eyepiece}=1+\frac{d}{f_{eyepiece}}=1+\frac{25}{5}=6$
From here, magnification by the objective lens :
$m_{objective}=\frac{30}{6}=5$ since ( $m_{objective}*m_{eyepiece}=m_{total}$ )
$m_{objective}=-\frac{v}{u}=5$
$v=-5u$
According to the lens formula:
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{1.25}=\frac{1}{-5u}-\frac{1}{u}$
from here,
$u = -1.5cm$
Hence object must be 1.5 cm away from the objective lens.
$v= -mu=(-1.5)(5)=7.5$
Now for the eyepiece lens:
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\frac{1}{5}=\frac{1}{-25}-\frac{1}{u}$
$\frac{1}{u}=-\frac{6}{25}$
$u = -4.17 cm$
Hence, the object is 4.17 cm away from the eyepiece lens.
The separation between objective and eyepiece lens
$u_{eyepiece} +v_{objectivve}=4.17 + 5.7 = 11.67 cm$
Answer:
Given,
the focal length of the objective lens $f_{objective}=140cm$
the focal length of the eyepiece lens $f_{eyepiece}=5cm$
Normally, the least distance of vision = 25cm
Now,
As we know, magnifying power:
$m = \frac{f_{objective}}{f_{eyepiece}}=\frac{140}{5}=28$
Hence magnifying power is 28.
Answer:
Given,
the focal length of the objective lens $f_{objective}=140cm$
the focal length of the eyepiece lens $f_{eyepiece}=5cm$
Normally, the least distance of vision = 25cm
Now,
As we know magnifying power when the image is at d = 25 cm is
$m=\frac{f_{objective}}{f_{eyepiece}}(1+\frac{f_{eyepiece}}{d})=\frac{140}{5}(1+\frac{5}{25}) = 33.6$
Hence, magnification, in this case, is 33.6.
Answer:
a) Given,
focal length of the objective lens = $f_{objective}$ = 140cm
focal length of the eyepiece lens = $f_{eyepiece}$ = 5 cm
The separation between the objective lens and eyepiece lens is given by:
$f_{eyepiece}+f_{objective}=140+5=145cm$
Hence, under normal adjustment separation between the two lenses of the telescope is 145 cm.
Answer:
Given,
focal length of the objectove lens = $f_{objective}$ = 140cm
focal length of the eyepiece lens = $f_{eyepiece}$ = 5 cm
Height of tower $h_{tower}$ = 100m
Distance of object which is acting like an object $u$ = 3km = 3000m.
The angle subtended by the tower at the telescope
$tan\theta=\frac{h_{tower}}{u}=\frac{100}{3000}=\frac{1}{30}$
Now, let the height of the image of the tower by the objective lens be $h_{image}$.
angle made by the image of the objective lens :
$tan\theta'=\frac{h_{image}}{f_{objective}}=\frac{h_{image}}{140}$
Since both the angles are the same, we have,
$tan\theta=tan\theta'$
$\frac{1}{30}=\frac{h_{image}}{140}$
$h_{image}= \frac{140}{30}=4.7cm$
Hence, the height of the image of the tower formed by the objective lens is 4.7 cm.
Q 9.28 (c) What is the height of the final image of the tower if it is formed at 25cm
Answer:
Given that the image is formed at a distance $d$ = 25cm
As we know, the magnification of the eyepiece lens is given by :
$m=1+\frac{d}{f_{eyepiece}}$
$m=1+\frac{25}{5}=6$
Now,
Height of the final image is given by :
$h_{image}= mh_{object}= 6*4.7 = 28.2cm$
Therefore, the height of the final image will be 28.2 cm
Answer:
Given,
Distance between the objective mirror and secondary mirror $d= 20mm$
The radius of curvature of the Objective Mirror
$R_{objective}=220mm$
So the focal length of the objective mirror
$f_{objective}=\frac{220}{2}=110mm$
The radius of curvature of the secondary mirror
$R_{secondary}=140mm$
So, the focal length of the secondary mirror
$f_{secondary}=\frac{140}{2}=70mm$
The image of an object which is placed at infinity, in the objective mirror, will behave like a virtual object for the secondary mirror.
So, the virtual object distance for the secondary mirror
$u_{secondary}=f_{objective}-d=110-20=90mm$
Now, applying the mirror formula in the secondary mirror:
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
$\frac{1}{v}=\frac{1}{70}-\frac{1}{90}$
$v=315mm$
Answer:
Given
Angle of deflection $\delta = 3.5^0$
The distance of the screen from the mirror is $D=1.5m$
The reflected rays will be deflected by twice the angle of deviation, that is
$2\delta =3.5*2=7^0$
Now from the figure, it can be seen that
$tan2\delta =\frac{d}{1.5}$
$d=1.5*\tan2\delta = 2*\tan7=0.184m=18.4cm$
Hence displacement of the reflected spot of the light is $18.4cm$.
Answer:
Given
The focal length of the convex lens $f_{convex}=30cm$
Here, the liquid is acting like a mirror, so
the focal length of the liquid $=f_{liquid}$
the focal length of the system(convex + liquid) $f_{system}=45cm$
The equivalent focal length when two optical systems are in contact
$\frac{1}{f_{system}}=\frac{1}{f_{convex}}+\frac{1}{f_{liquid}}$
$\frac{1}{f_{liquid}}=\frac{1}{f_{system}}-\frac{1}{f_{convex}}$
$\frac{1}{f_{liquid}}=\frac{1}{45}-\frac{1}{30}=-\frac{1}{90}$
$f_{liquid}=-90cm$
Now, let us assume the refractive index of the lens to be $\mu _{lens}$
The radius of curvature is $R$ and $-R$.
As we know,
$\frac{1}{f_{convex}}=(\mu _{lens}-1)\left ( \frac{1}{R}-\frac{1}{-R} \right )$
$\frac{1}{f_{convex}}=(\mu _{lens}-1)\frac{2}{R}$
$R=2(\mu _{lens-1})f_{convex}=2(1.5-1)30=30cm$
Now, let the refractive index of the liquid be $\mu _{liquid}$
The radius of curvature of the liquid in the plane mirror side = infinite
Radius of curvature of liquid in lens side R = -30cm
As we know,
$\frac{1}{f_{liquid}}=(\mu_{liquid-1})\left ( \frac{1}{R}-\frac{1}{infinite} \right )$
$-\frac{1}{90}=(\mu_{liquid}-1)(\frac{1}{30})$
$\mu_{liquid}= 1+\frac{1}{3}$
$\mu_{liquid}= 1.33$
Therefore, the refractive index of the liquid is 1.33.
Also Read
Class 12 Physics Chapter 9 - Ray Optics and Optical Instruments: Higher Order Thinking Skills (HOTS) Questions
Class 12 Physics Chapter 9 - Ray Optics and Optical Instruments HOTS questions are designed to test a deep understanding of concepts like refraction, total internal reflection, and optical instruments. These advanced problems sharpen problem-solving skills and are highly useful for board exams as well as JEE and NEET preparation.
Q.1 A convex lens made of glass (refractive index $=$ 1.5) has a focal length of 24 cm in air. When it is totally immersed in water (refractive index $=1.33$ ), its focal length changes to
Answer:
$
\begin{aligned}
& \frac{1}{8}=\left(\frac{\mu_{\ell}}{\mu_{\mathrm{s}}}-1\right)\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right] \\
& \frac{1}{24}=(1.5-1)\left[\frac{2}{\mathrm{R}}\right] ---(i)\\
& \frac{1}{\mathrm{f}^{\prime}}=\left(\frac{1.5}{1.33}-1\right)\left(\frac{2}{\mathrm{R}}\right) \\
& \frac{1}{\mathrm{f}^{\prime}}=\left(\frac{1.5 \times 3}{4}-1\right) \frac{2}{\mathrm{R}}---(ii)
\end{aligned}
$
(i) divided by (ii)
$\begin{aligned} & \frac{\mathrm{f}^{\prime}}{24}=4 \\ & \mathrm{f}^{\prime}=96 \mathrm{~cm} \\ & \text { }\end{aligned}$
Q.2 An Ice cube has a bubble inside. When viewed from one side, the apparent distance of the bubble is $24 \mathrm{~cm}$. When viewed from the opposite side, the apparent distance of the bubble is observed as $2 \mathrm{~cm}$. If the side of the ice cube is $48 \mathrm{~cm}$. The refractive index of the cube is (answer should be the nearest integer)
Answer:
let side of the cube is $x$
refractive index $=n$
$
n=\frac{\text { true depth }}{\text { App depth }}
$
When viewing the bubble from one side, the depth is $x$, and the apparent depth is $24 \mathrm{~cm}$.
$ n=\frac{x}{24} ......(i) $
When viewing the bubble from the opposite side, the true depth is $48-x$
(since the side of the cube is $48 \mathrm{cm})$. and the apparent depth in $2 \mathrm{~cm}$.
$n=\frac{48-x}{2}.....(ii)$
from i and ii
$
\begin{aligned}
\frac{x}{24} & =\frac{48-x}{2} \\
\frac{x}{12} & =48-x \\
x & =576-12 x \\
13 x & =576 \\
x & =\frac{576}{13} \\
x & =44.3 \approx 44
\end{aligned}
$
So,
$
n=\frac{44}{12}=\frac{11}{3}=3.67
$
The nearest integer is 4
so n=4
Q.3 A point object is moving with a speed of v before an arrangement of two mirrors as shown in the figure.
Find the velocity of image in mirror $\mathrm{M}_1$ with respect to image in mirror $\mathrm{M}_2$
Answer:
Velocity of image, $\mathrm{v}_{\mathrm{r}}=\sqrt{\mathrm{v}^2+\mathrm{v}^2-2 \mathrm{v} \cdot \mathrm{v} \cdot \cos \theta}=2 \mathrm{v} \sin (\theta / 2)$
Q.4 A metal plate is lying at the bottom of a tank full of a transparent liquid. The height of the tank is 100cm, but the plate appears to be at 45 cm above the bottom. The refractive index of the liquid is:
Answer:
Real depth of plate, $\mathrm{H}=100 \mathrm{~cm}$
The apparent depth of the plate, $\mathrm{h}=100-45=55 \mathrm{~cm}$
$\therefore$ The refractive index of the fluid $=\frac{\mathrm{H}}{\mathrm{h}}=\frac{100}{55}=1.81$
Q. 5 Two lenses are placed in contact with each other, and the focal length of the combination is 80cm. If the focal length of one is 20cm, then the power of the other will be:
Answer:
The focal length of the combination,
$
\begin{aligned}
& \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \Rightarrow \frac{1}{80}=\frac{1}{20}+\frac{1}{\mathrm{f}_2} \\
& \therefore \mathrm{f}_2=-\frac{80}{3} \mathrm{~cm}
\end{aligned}
$
$\therefore \quad$ Power of second lens $\mathrm{P}^{\prime}=\frac{100}{\mathrm{f}_2}=\frac{100}{(-80 / 3)}=-3.75 \mathrm{D}$
Ray Optics and Optical Instruments NCERT Solutions: Topics
Class 12 Physics Chapter 9 - Ray Optics and Optical Instruments deals with reflection, refraction, total internal reflection, prism, formulae of lens and mirror, the working of microscopes and telescopes. These concepts act as the basis of the comprehension of image formation and optical devices in the board exams, as well as in the competitive exams.
9.1 Introduction
9.2 Reflection of light by spherical mirrors
9.2.1 Sign convention
9.2.2 Focal length of spherical mirrors
9.2.3 The mirror equation
9.3 Refraction
9.4 Total internal reflection
9.4.1 Total internal reflection in nature and its technological applications
9.5 Refraction at spherical surfaces and by lenses
9.5.1 Refraction at a spherical surface
9.5.2 Refraction by a lens
9.5.3 Power of a lens
9.5.4 Combination of thin lenses in contact
9.6 Refraction through a prism
9.7 Optical instruments
9.7.1 The microscope
9.7.2 Telescope
CBSE Class 12th Syllabus: Subjects & Chapters
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Ray Optics and Optical Instruments NCERT Solution: Important Formulas
The significant equations of Class 12 Physics NCERT Chapter 9 Ray Optics and Optical Instruments will serve as a quick reference guide to solve numerical problems on reflection and refraction, as well as the functioning of optical instruments. These formulas play an important role in both board exams and in competitive exams such as JEE and NEET, in helping students revise faster and be able to apply the concepts learned correctly in solving problems.
1. Mirror Formula
$
\frac{1}{f}=\frac{1}{v}+\frac{1}{u}
$
Where $f=$ focal length, $\underline{u}=$ object distance, $\underline{v}=$ image distance.
2. Magnification by Mirror
$
M=\frac{h_i}{h_o}=-\frac{v}{u}
$
3. Lens Formula
$
\frac{1}{f}=\frac{1}{v}-\frac{1}{u}
$
4. Magnification by Lens
$
M=\frac{h_i}{h_o}=\frac{v}{u}
$
5. Lens Maker's Formula
$
\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
$
Where $n=$ refractive index, $R_1, R_2 \equiv$ radii of curvature.
6. Refraction at Spherical Surface
$
\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}
$
7. Lateral Shift in a Slab
$
\Delta x=\frac{t \sin (i-r)}{\cos r}
$
Where ${t}=$ thickness of slab, ${i}=$ angle of incidence, ${r}=$ angle of refraction.
8. Prism Formula (Deviation)
$
\delta=\left(i_1+i_2\right)-A
$
At minimum deviation:
$
\mu=\frac{\sin \left(\frac{A+D_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}
$
9. Magnifying Power of a Simple Microscope
$
M=1+\frac{D}{f}
$
Where $D=$ least distance of distinct vision ( $\sim 25 \mathrm{~cm}$ ), ${f=}$ focal length of lens.
10. Magnifying Power of Compound Microscope
$
M=\frac{v_o}{u_o} \times\left(1+\frac{D}{f_e}\right)
$
Where $v_o, u_o=$ image $\&$ object distance for objective, $f_e=$ focal length of eyepiece.
11. Magnifying Power of Telescope (Normal Adjustment)|
$
M=-\frac{f_0}{f_e}
$
Where $f_o=$ focal length of objective, $f_e=$ focal length of eyepiece.
12. Resolving Power of Microscope
$
d=\frac{1.22 \lambda}{2 \mu \sin \theta}
$
13. Resolving Power of Telescope
$
d \theta=\frac{1.22 \lambda}{D}
$
Where $D=$ diameter of objective lens.
How Can NCERT Solutions for Class 12 Physics Chapter 9 Help in Exam Preparation?
Ray Optics and Optical Instruments class 12 question answers are an essential resource for students preparing for their board and competitive exams. Some important concepts covered in this chapter are the reflection, refraction, how mirrors and lenses form images and the mechanics of how optical instruments like telescopes and microscopes operate. The Class 12 physics Ray Optics and Optical Instruments question answers enable students to grasp these concepts in stages with easy-to-read and well-spaced diagrams that enable one to learn more easily and in a more visual manner.
Through practising the solved examples and the exercise questions regularly, students have an opportunity to improve their numerical-solving abilities and understanding of the concepts. Significant derivations and real-life practice are also found in the Ray Optics and Optical Instruments Class 12 NCERT Solutions, and they assist students in correlating theory with practical knowledge. In addition, learning using the Ray Optics and Optical Instruments Class 12 Questions and Answers gives students an assurance that they remain within the CBSE marking scheme, revise effectively for exams, and have the confidence of answering both conceptual and application-based questions with precision.
Approach to Solve Questions of Class 12 Physics Chapter 9 - Ray Optics and Optical Instruments
It is essential to have a good understanding of reflection, refraction, lens formula, mirror equation, and optical instruments to solve questions on Chapter 12 Physics Class 9 Ray Optics and Optical Instruments. Students will have to use sign conventions, light ray geometry and formulae with care and caution to ensure that problems are solved accurately and precisely, both theoretically and numerically. The step-by-step approach is systematic and makes everything clear and precise.
- Get a Firm Idea of the Concept:
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Determine what problem is on reflection, refraction, or total internal reflection (TIR) and optical instruments.
-
Learn the corresponding laws, e.g. Snell's law, laws of reflection, or lens/mirror formula.
-
Draw a Proper Ray Diagram
-
Image formation can be visualised using neat ray diagrams.
-
Make sure to mark incident rays and refracted rays, as well as normals, so that they are not confused.
-
Use Sign Convention (New Cartesian Convention):
-
In the case of mirrors and lenses, obey the sign conventions, i.e. distances which are measured against the direction of incident light are negative, and those in the direction of incident rays are positive.
-
Use Standard Formulas Accurately:
Mirror formula: $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
Lens formula: $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
Magnification in lens: $M=\frac{h^{\prime}}{h}=\frac{v}{u}$
-
Break Down Complex Problems:
-
In the case of many refractions (such as a lens system or prism), solve step-by-step.
-
Treat air-glass (glass-air) interfaces individually, acquire the results, and combine.
-
Check Units and Consistency:
-
Before proceeding to substitution, all measurements should be converted to one common unit (cm, m, etc.).
-
Check your calculation again with regard to sign errors.
-
Relate to Real-Life Applications:
Connect numerical results with practical understanding (e.g., why convex lenses form real images, how prisms disperse light, etc.).
Importance of NCERT Solutions for Class 12 Physics Chapter 9: Ray Optics
Class 12 Physics chapter 9 Ray Optics question answers are essential for understanding the behaviour of light when it interacts with mirrors, lenses, and optical instruments. The Class 12 Physics NCERT Chapter 9 Ray Optics and Optical Instruments integrates theoretical knowledge with problem-solving using graphs and numericals, and so proper guidance is essential in this case.
- Helps students have a clear understanding of the reflections and refractions of light through the use of mirrors and lenses.
- Describes the most important Snell's law, mirror formula and lens formula in an organised manner.
- Develops good abilities to make the correct ray diagrams, which is crucial when it comes to scoring in board exams.
- Plays a key role in CBSE board exams, where ray diagrams, derivations, and numericals are frequently asked.
- Highly useful for competitive exams like JEE and NEET, especially questions involving image formation and optical instruments.
- Enhances the capability of solving mathematical problems by step-by-step solutions of questions on mirrors and lenses
- Makes students know how optical devices like microscopes and telescopes work.
- Helps to use sign conventions and formulae correctly, eliminating typical examination errors.
What Students Learn from NCERT Solutions for Class 11 Physics Chapter 9: Ray Optics?
The Class 12 Physics NCERT Chapter 9 Ray Optics and Optical Instruments presents the students with the geometrical method of the study of light and its interaction with flat, reflective and refractive surfaces. By using NCERT Ray Optics and Optical Instruments Class 12 Solutions, students are able to clearly understand the formation of the image, the law of optics, and diagram-based concepts that are vital in the theory and numerical questions in the examinations.
- Know the fundamentals of laws of reflection and refraction, and how they can be applied to mirrors and lenses.
- Practice of using the mirror formula, the lens formula and the relation of magnification in numerical problems.
- Learn to create sharp and precise ray diagrams to show the formation of images.
- Learn about refraction using prisms, deviation and dispersion of light.
- Study the principles of working of optical instruments like microscopes and telescopes.
- Develop a good background in more advanced topicssuch as wave optics and optical instruments.
What Extra Should Students Study Beyond NCERT for JEE/NEET?
Beyond NCERT, students preparing for JEE/NEET should focus on advanced concepts of Ray Optics like derivations of lens-maker's formula, optical path, dispersion and achromatic combination of lenses, resolving power of optical instruments, and advanced problems on prism and total internal reflection. These additional concepts strengthen problem-solving skills and help tackle tricky application-based questions in competitive exams.
NCERT Solutions for Class 12 Physics: Chapter-Wise
NCERT Class 12 Physics Solutions provide chapter-wise, detailed answers to all textbook exercises, making it easier for students to grasp tough concepts. These step-by-step solutions are designed as per the latest CBSE syllabus and are highly useful for both board exams and competitive exams like JEE & NEET. With chapter-wise links, students can directly access the solutions and download the PDFs for quick and effective revision.
NCERT Books and NCERT Syllabus
NCERT Solutions Subject-wise
NCERT Solutions subject-wise links provide easy access to chapter-wise answers for all major subjects in one place. These solutions help students understand concepts clearly, practise textbook questions effectively, and prepare confidently for exams
NCERT Exemplar Class 12 Solutions
NCERT Exemplar Solutions for Class 12 provide advanced and application-based questions to strengthen conceptual understanding. These subject-wise links help students practise higher-level problems and prepare effectively for board exams as well as competitive exams like JEE and NEET.
Frequently Asked Questions (FAQs)
Q: What is the weightage of the ray optics for CBSE board exam
A:
From the NCERT chapter ray optics, 6 to 9 marks questions are asked for CBSE board exam
Q: What does the NCERT Solutions for Class 12 Physics Chapter 9 PDF include?
A:
The NCERT Solutions for Class 12 Physics Chapter 9 PDF includes step-by-step answers to textbook questions, explanations of key concepts, and diagrams to help you understand Ray Optics and Optical Instruments better.
Q: How useful is the Ray Optics Class 12 NCERT PDF for board exam preparation?
A:
The Ray Optics Class 12 NCERT PDF is very helpful for board exams as it is based on the CBSE syllabus and includes solved exercises, conceptual clarity, and practice problems aligned with exam patterns.
Q: Where can I find the class 12 physics chapter 9 ncert solutions exercise-by-exercise answers?
A:
The right study materials should be chosen by Class 12 students in order to encourage efficient textbook problem-solving. Finding the best reference book from the many available on the market demands for a lot of patience. In Careers360 the answers to the chapter- and exercise-specific issues are given in PDF format. Students can use it to quickly dispel their doubts while working through challenges.
Q: What are the main differences between ray optics and wave optics, as described in the NCERT Solutions for Class 12?
A:
Ray optics uses laws like reflection and refraction to focus on the creation of geometric images by treating light as straight lines, or rays. In order to explain phenomena like interference, diffraction, and polarization, wave optics views light as a wave. NCERT Solutions distinguishes between two applications: ray optics is utilized for image location and instrument design, but wave optics describes patterns and color effects that are not predicted by rays.
Q: What circumstances could lead to the failure of Ray Optics' ray approximation?
A:
- When the size of an item or aperture is similar to the wavelength of light (diffraction/wave optics phenomena)
- When rays approach the edge of the lens or mirror, they produce a large amount of spherical aberration.
- At extremely precise or invisible wavelengths
NCERT advises students to be careful of these restrictions, particularly while answering concept and HOTS questions.
Q: In CBSE board exams, which Ray Optics and Optical Instruments topics are most commonly asked?
A:
- Mirror and lens formula derivation and application
- The applications of total internal reflection, such as optical fibers
- Numericals on optical devices (telescope, compound microscope)
- Ray diagrams for mirror image formation and concave/convex lenses
- Applications of sign conventions in numerical
These topics are essential for exam preparation since they are given priority in NCERT Solutions and board examinations.
Q: What are the common mistakes students make in Ray Optics?
A:
Mistakes that happen frequently are: wrong sign convention, wrong ray diagrams, and wrong location of images.
Q: What is Snell's law of refraction?
A:
According to Snell law, the ratio of the sine of the angle of incidence to sine of the angle of refraction remains constant with respect to a given pair of media.
Q: Which Topics Have the Highest Weightage in NEET?
A:
- Lens Formula & Magnification (Frequently Asked).
- Total Internal Reflection (Applications like optical fibres).
- Prism-related problems (Minimum deviation, angle of prism).
- Microscope & Telescope (Magnifying power, resolving power).
- Human Eye Defects (Myopia, Hypermetropia, Presbyopia).
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Questions related to CBSE Class 12th
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Hello
You will be able to download the CBSE Previous Year Board Question Papers from our official website, careers360, by using the link given below.
https://school.careers360.com/boards/cbse/cbse-previous-year-question-papers
I hope this information helps you.
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Hello
You will be able to download the CBSE Pre-Board Class 12 Question Paper 2025-26 from our official website by using the link which is given below.
https://school.careers360.com/boards/cbse/cbse-pre-board-class-12-question-paper-2025-26
I hope this information helps you.
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Hello,
Yes, it's completely fine to skip this year's 12th board exams and give them next year as a reporter or private candidate, allowing you to prepare better; the process involves contacting your current school or board to register as a private candidate or for improvement exams during the specified
HELLO,
Yes i am giving you the link below through which you will be able to download the Class 12th Maths Book PDF
Here is the link :- https://school.careers360.com/ncert/ncert-book-for-class-12-maths
Hope this will help you!
Hello,
Here is your Final Date Sheet Class 12 CBSE Board 2026 . I am providing you the link. Kindly open and check it out.
https://school.careers360.com/boards/cbse/cbse-class-12-date-sheet-2026
I hope it will help you. For any further query please let me know.
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Popular CBSE Class 12th Questions
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
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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
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A particle is projected at 600 to the horizontal with a kinetic energy . The kinetic energy at the highest point
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In the reaction,
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How many moles of magnesium phosphate, 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 |
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