NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics 2021

NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

Edited By Vishal kumar | Updated on Sep 11, 2023 01:26 PM IST | #CBSE Class 12th

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NCERT solutions for class 12 Physics chapter 10 Wave Optics - Students appearing in the Class 12 board exams must go through this Solutions for Class 12. These NCERT solutions for Class 12 Physics chapter 10 Wave Optics gives you an insight into how to deal with the questions asked in the NCERT books for Class 12 Physics. NCERT Class 12 Physics solutions chapter 10 Wave Optics comprise of the step by step solutions in an easy-to-understand language. The NCERT solutions for Class 12 Physics chapter 10 Wave Optics are helpful in academics as well as in competitive exams. Read further to know the Class 12 Physics Chapter 10 NCERT solutions.

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NCERT Solutions for Class 12 Physics Chapter 10 – Download Free PDF
NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics
NCERT solutions for Class 12 Pysics chapter 10 Wave Optics additional exercises:
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NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

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NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

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NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics: Exercises Solution

Q10.1 (a) Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected. Refractive index of water is 1.33.

Answer:

Given a monochromatic light has a wavelength of $589 nm\ or\ 589\times 10^{-9}m$ .

And as we know the speed of the light in air is, $c = 3\times 10^{8} m/s$ .

Refractive index of water, $\mu = 1.33$

Therefore we have now,

The ray is incident on the medium, and then it gets reflected back to the same medium therefore, the wavelength, speed, and frequency will be the same as that of the incident ray.

We have the frequency of light is given by the relation,

$v = \frac{c}{\lambda}$

$= \frac{3\times 10^8 m/s}{589\times 10^{-9}}$

$= 5.09\times 10^{14}Hertz$ $= 5.09\times 10^{14}Hertz$

Therefore the speed, frequency, and the wavelength of the reflected light are $3\times 10^{8} m/s$ , $5.09\times 10^{14}Hertz$ , and $589 nm$ respectively.

Q10.1 (b) Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (b) refracted light? Refractive index of water is 1.33.

Answer:

Given,

The wavelength of the monochromatic light $\lambda =589nm=589*10^{-9}m$

Refractive index of the water $\mu_{water}=1.33$

b)In the case of refraction, speed and wavelength will change as the medium is changing however, the frequency will remain the same because it doesn't change when the medium is changed,

so, frequency $f=5.09 * 10^{14}Hz$

Speed of rays: speed of rays in water with refractive index 1.33 is

$v=\frac{c}{\mu _{water}}=\frac{3*10^8}{1.33}=2.26*10^8m/s$

Now, the Wavelength of light

$\lambda =\frac{v}{f}=\frac{2.26*10^8}{5.09*10^{14}}=444.01*10^{-9}m$

Hence the wavelength of light is 444.01nm, frequency is $5.09 * 10^{14}Hz$ and speed is $2.26*10^8m/s$ .

Q10.2 (a) What is the shape of the wavefront in each of the following cases:

(a) Light diverging from a point source.

Answer:

The shape of wavefront when light is diverging from a point source is Spherical since light travels in all direction.

Q10.2 (b) What is the shape of the wavefront in each of the following cases:

(b) Light emerging out of a convex lens when a point source is placed at its focus.

Answer:

The shape of light emerging out of a convex lens when a point source is placed at its focus is Parallel .when light rays come from infinity parallelly, they intersect at focus of convex lens and hence when light is emerging from the focus, the rays will get parallel to each other after coming out of the convex lens, because path of light rays are reversible.

Q10.2 (c) What is the shape of the wavefront in each of the following cases:

Answer:

The portion of the wavefront of the light from the distant star which is seen from earth is plane since a small area of a large sphere will nearly look like a plane.

Q10.3 (a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is $3.0 \times 10^8 m s^{-1}$ )

Answer:

Given,

Refractive index of the glass $\mu _{glass}$ = 1.5

Speed of light in vaccum $c=3*10^8m/s$

Now,

As we know,

Refractive index of a medium

$\mu _{medium }=\frac{c}{v_{medium}}$ where ${v_{medium}}$ is the speed of light in that medium.

so from here,

${v_{glass}}=\frac{c}{\mu _{glass}}=\frac{3*10^8}{1.5}=2*10^8m/s$

Hence the speed of light in water is $2*10^8m/s$

Q10.3 (b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?

Answer:

No, the speed of light in glass is not independent of the colour. The colour of the light does influence the refractive index and speed of light in the medium. The refractive index of the violet light is greater than the refractive index of red light and hence red component of the white light travels faster in the glass than the red component of the light.

$v=\frac{c}{\mu }$

the more the refractive index, the lesser the speed.

Q10.4 In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

Answer:

Given,

Distance between screen and slit $D=1.4m$

Distance between slits $d=0.28mm=0.28*10^{-3}m$

Distance between central and fourth bright fringe $u=1.2cm=1.2*10^{-2}m$

Now,

as we know, the distance between two fringes in a constructive interference is given by

$u=n\lambda \frac{D}{d}$

where $n=$ order of fringe (which is 4 here) and $\lambda$ is the wavelength of light we are using.

so from here,

$\lambda = \frac{ud}{nD}=\frac{1.2*10^{-2}*0.28*10^{-3}}{4*1.4}=6*10^{-7}m$

Hence wavelength os the light is 600nm

Q10.5 In Young’s double-slit experiment using monochromatic light of wavelength $\lambda$ , the intensity of light at a point on the screen where path difference is $\lambda$ , is K units. What is the intensity of light at a point where path difference is $\frac{\lambda}{3}$ ?

Answer:

Given, in youngs double-slit experiment.

the wavelength of monochromatic light = $\lambda$

The intensity of light when the path difference is $\lambda$ = K

Now,

As we know,

The phase difference $\phi$ is given by

$\phi =\frac{2\pi }{\lambda }(PathDiffernce)$

also

Total Intensity

$I=I_1+I_2+2\sqrt{I_1I_2}cos\phi$

Let $I_1=I_2=I_0$

Now, when path difference is $\lambda$

the phase difference angle

$\phi=\frac{2\pi }{\lambda}*\lambda=2\pi$

so,

$I_0+I_0+2\sqrt{I_0I_0}cos2\pi=K$

$K=4I_0$

Now, when path difference is $\frac{\lambda }{3}$

$\phi=\frac{2\pi }{\lambda}*\frac{\lambda}{3}=\frac{2\pi}{3}$

Intensity of light

$K'=I_0+I_0+2\sqrt{I_0I_0}cos\frac{2\pi}{3}$

$K'=I_0$

Now comparing intensity at both cases

$\frac{K'}{K}=\frac{I_0}{4I_0}=\frac{1}{4}$

$K'=\frac{K}{4}$

Hence intensity will reduce to one-fourth of initial when path difference changes from $\lambda$ to $\frac{\lambda}{3}$ .

Q10.6 A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.

(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.

Answer:

Given,

The wavelength of one light beam :

$\lambda_1=650nm=650*10^{-9}m$

The wavelength of another Light beam

$\lambda_2=520nm=520*10^{-9}m$

Let, the distance between the two-slit be $d$ and distance between slit and screen is $D$

Now,

As we know, the distance $x$ of nth bright fringe from central maxima is given by

$x=n\lambda \frac{D}{d}$

so for 3rd fringe,

$n=3$

$x=n\lambda_1 \frac{D}{d}=3*650*10^{-9}*\frac{D}{d}=1950\frac{D}{d}nm$

Hence distance of 3rd fringe from central maxima is $1950\frac{D}{d}nm$ . Here value D and d are not given in the question.

Q10.6 A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.

(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?

Answer:

Let the nth bright fringe due to wavelength, $\lambda_2$ and (n − 1) ^th bright fringe due to wavelength $\lambda_1$ coincide on the screen. We can equate the conditions for bright fringes as:

$n\lambda_2=(n-1)\lambda _1$

$520n=650n-650$

$650=130n$

$n=5$

Hence, the least distance from the central maximum can be obtained by the relation:

$X=n\frac{\lambda_2D}{d}=5*\frac{520*10^{-9}*D}{d}$

Hence we can find out X if D and d are given.

Q10.7 In a double-slit experiment the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water? Take refractive index of water to be 4/3.

Answer:

Given

The angular width of the fringe when the medium is air

$\theta _{air}=0.2^0$

The distance of the screen from the slit $D = 1m$

The wavelength of light we are using $\lambda=600nm=600*10^{-9}m$

Refractive index of water $\mu_{water}=4/3$

let angular width of fringe when the medium is water $\theta _{water}$

Now, as we know the angular width is given by

$\theta =\frac{\lambda }{d}$

so,

$d=\frac{\lambda _{air}}{\theta _{air}}=\frac{\lambda _{water}}{\theta _{water}}$

$d=\frac{\lambda _{air}}{\lambda _{water}}=\frac{\theta _{air}}{\theta _{water}}=\mu$

From here

$\theta _{water}=\frac{\theta _{air}}{\mu _{water}}=\frac{3}{4}0.2^0=0.15^0$

Hence angular width of the fringe in the water is $0.15^0$ .

Q10.8 What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)

Answer:

Given,

Refractive index of glass $\mu_{glass}=1.5$

Now as we know,

$tan\theta =\mu$

where $\theta$ is the polarizing angle, also called the Brewster angle.and $\mu$ is the refractive index.

so from here

$\theta =tan^{-1}(\mu _{glass})=tan^{-1}1.5=56.31^0$

Hence Brewster angle is $56.31^0$ .

Q10.9 Light of wavelength 5000 Å falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light? For what angle of incidence is the reflected ray normal to the incident ray?

Answer:

Given,

Wavelength of light $\lambda =5000\AA =5000*10^{-10}m$

Speed of light $c=3*10^8$

Now,

Wavelength and frequency will be the same when the ray is reflected.

Frequency of reflected light

$f=\frac{c}{\lambda }=\frac{3*10^{8}}{5000*10^{-10}}=6*10^{14}Hz$

Hence wavelength and frequency of light is $5000*10^{-10}m$ and $6*10^{14}Hz$ respectively.

Now,

as per the law of reflection, angle of incidence $i$ is always equal to angle of reflection $r$ .

$i=r$

Now, when the reflected ray is perpendicular with incidence ray,

$i+r=90$

$i+i=90$

$2i=90$

$i=45$

Hence the angle of incidence is 45 for this condition.

Q10.10 Estimate the distance for which ray optics is good approximation for an aperture of 4 mm and wavelength 400 nm.

Answer:

Given

Aperture $a=4mm=4*10^{-3}m$

Wavelength of light $\lambda =400nm=400*10^{-9}m$

Now,

Distance for which ray optics is a good approximation also called Fresnel's distance:

$Z_f=\frac{a^2}{\lambda }=\frac{(4*10^{-3})^2}{400*10^{-9}}=40m$

Hence distance for which ray optics is a good approximation is 40m.

NCERT solutions for Class 12 Pysics chapter 10 Wave Optics additional exercises:

Q10.11 The 6563 Å Hα line emitted by hydrogen in a star is found to be red-shifted by 15 Å. Estimate the speed with which the star is receding from the Earth.

Answer:

Given,

wavelength Hα line emitted by hydrogen:

$\lambda=6563*10^{-10}m$

star is red-shifted by

$\lambda'-\lambda=15*10^{-10}m$

let velocity of the star be $v$

Now,

as we know,

$\lambda'-\lambda=\frac{v}{c}*\lambda$

from here

$\lambda'-\lambda\frac{1}{{c}*\lambda }={v}$

$v=\frac{3*10^8*15*10^{-10}}{6563*10^{-10}}=6.87*10^5m/s$

Hence speed at which star is receding away is $6.87*10^5m/s$

Q10.12 Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water? If not, which alternative picture of light is consistent with experiment?

Answer:

According to corpuscular theory, when corpuscle of the light goes from rare medium to denser medium, the component of their velocity along the surface of the interface remains the same.

So we can write

$v_1sini=v_2sinr$

$\frac{v_2}{v_1}=\frac{sini}{sinr}=\mu$

As $\mu>1$ ,

$v_2>v_1$

That is light should be faster in the dense medium than in rare medium.this is the opposite of what we see experimentally.

Huygens wave theory predicts that light is faster in a rare medium which matches with our experiments observation.

Q10.13 You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.

Answer:

Let an object M is placed in front of a plane mirror AB at a distance r .

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A circle is drawn from the centre, such that it just touches the plane mirror at point P. according to the Huygens’ principle, AB is the wavefront of the incident light .

If the mirror is absent then a similar wavefront A'B' would form behind M at a distance r.

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A'B' can be considered as a virtual reflected ray for the plane mirror. Hence a point object placed in front of the plane mirror produces an image at the same distance as the object.

Q10.14 (a) Let us list some of the factors, which could possibly influence the
speed of wave propagation:
(i) nature of the source.
ii) direction of propagation.
(iii) motion of the source and/or observer.

(iv) wavelength.
(v) intensity of the wave.
On which of these factors, if any, does
(a) the speed of light in vacuum depend?

Answer:

The speed of light in a vacuum is constant and independent of anything according to Einstein's theory of relativity.

Question:

Q10.14 (b) Let us list some of the factors, which could possibly influence the speed of wave propagation:
(i) nature of the source.
(ii) direction of propagation.
(iii) motion of the source and/or observer.
(iv) wavelength.
(v) intensity of the wave.
On which of these factors, if any, does

(b) the speed of light in a medium (say, glass or water), depend?

Answer:

The speed of light in any medium depends upon the wavelength of the light and does not depends on the nature of the source, direction of propagation, the motion of the source and/or observer, and intensity of the wave.

Q10.15 For sound waves, the Doppler formula for frequency shift differs slightly between the two situations: (i) source at rest; observer moving, and (ii) source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?

Answer:

The sound wave requires a medium for propagation.so, even though both given situation may relate to the same relative motion, they are not identical physically since, the motion of the observer, relative to the medium is different in two situations. Hence, we cannot expect the Doppler formula to be identical in both given cases.

When light waves are in a vacuum, there is clearly nothing to distinguish between two cases.

for light propagation in a medium, two situations are not identical for the same reason as in the case of sound waves.

Q10.16 In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between the two slits?

Answer:

Given,

Wavelength of light $\lambda$ = 600nm

Angular fringe width

$B_{\theta}=\frac{\lambda}{d}$

$d=\frac{\lambda}{B_{\theta}}=\frac{600*10^{-9}*180}{0.1*\pi }=3.44*10^{-4}m$

Hence spacing required between the two slits is $3.44*10^{-4}m$ .

Q10.17 (a) Answer the following questions:

In a single slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band?

Answer:

As we know,

width of the central diffraction band is given by

$2D\frac{\lambda}{d}$

where d is the width of the slit.

So when we double the width of the slit, the size of the central diffraction band reduces to half of its value. But, the light amplitude becomes double, which increase the intensity 4 times.

10.17 (b) Answer the following questions:

In what way is diffraction from each slit related to the interference pattern in a double-slit experiment?

Answer:

When we have a width in the order of $\lambda$ , the intensity of interference fringes in Young's double-slit experiment is modified by the diffraction pattern of each slit.

Q10.17 (c) Answer the following questions:

When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain why?

Answer:

A bright spot is seen at the centre of the shadow of the obstacle because wave diffracted from the edge of a circular obstacle interfere constructively at the centre of the shadow producing the bright spot.

Q10.17 (d) Answer the following questions:

Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily.

Answer:

The size of obstacle should be comparable to the wavelength for diffraction of waves by obstacles, through a large scale.

This comes from

$sin\theta =\frac{\lambda}{a}$

$sin\theta =\frac{10^{-7}}{10}=10^{-8}$

This implies $\theta\rightarrow 0$

it means the light goes almost unbent and hence student are unable to see each other.

Q10.17 (e) Answer the following questions:

Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?

Answer:

Typical size of the obstacle is much larger than the wavelength of light. Hence the diffraction effect is negligibly small. thus the assumption that light travels in a straight line can be safely used in day to day life.

Q10.18 Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects?

Answer:

Given,

Distance between two towers = 40km

size of aperture = $a=50m$

Now,

As we know

Fresnel's distance is equal to half of the distance between towers

$Z_f=\frac{40}{2}=20km$

Also from the formula:

$Z_f=\frac{a^2}{\lambda}=20m$

$\lambda=\frac{a^2}{Z_f}=\frac{50^2}{20*10^3}=12.5cm$

Hence this is the required longest wavelength of the radio wave, which can be sent in between the towers without considerable diffraction effect.

Q10.19 A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.

Answer:

Given

The distance of the screen from the slit, $D=1m$

The distance of the first minimum $X_1=2.5mm=10^{-3}=2.5*10^{-3}mm$

The wavelength of the light $\lambda=500nm=500*10^{-9}m$

Now,

As we know,

$X_n=n\frac{\lambda D}{d}$

$d=n\frac{\lambda D}{X_n}=1*\frac{500*10^{-9}*1}{2.5*10^{-3}}=2*10^{-4}m=0.2mm$

Hence, the width of the slit is 0.2 mm.

Q10.20 (a) Answer the following questions:
When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.

Answer:

When a low flying aircraft passes overhead, we notice slight shaking in pictures of the TV. This is because aircraft interferes with signals and reflects it. So the shaking we see is the interference of direct signal and reflected signal.

Q10.20 (b) Answer the following questions:

As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?

Answer:

The superposition principle comes from the linear character of the differential equation of wavemotion.that is if $x_1$ and $x_2$ are the solution of any wave equation, then linear combination of $x_1$ and $x_2$ is also the solution of the wave equation.

Q10.21 In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of $\frac{n\lambda}{a}$ . Justify this by suitably dividing the slit to bring out the cancellation.

Answer:

Let the width of the slit $b$ be divided into n equal parts so that

$b'=\frac{b}{n}$

$b=b'n$

Now,

$\theta=\frac{n\lambda}{b}=\frac{n\lambda}{b'n}=\frac{\lambda}{b'}$

At this angle, each slit will make the first diffraction minimum. therefore the resultant intensity for all the slits will be zero at the angle of $\frac{n\lambda}{b}$ .

Wave optics class 12 exercise solutions hold great significance for board exams, as well as competitive exams like JEE and NEET. This chapter's questions range from moderately challenging to complex, offering students the opportunity to earn substantial marks. By thoroughly understanding concepts related to interference, diffraction, and polarization, and practising with these solutions, students can confidently tackle questions and secure good marks in both board and competitive exams.

Chapter wise NCERT Solutions for class 12 physics

NCERT solutions for class 12 physics chapter 1 Electric Charges and Fields

NCERT solutions for class 12 physics chapter 2 Electrostatic Potential and Capacitance

NCERT solutions for class 12 physics chapter 3 Current Electricity

NCERT solutions for class 12 physics chapter 4 Moving Charges and Magnetism

NCERT solutions for class 12 physics chapter 5 Magnetism and Matter

NCERT solutions for class 12 physics chapter 6 Electromagnetic Induction

NCERT solutions for class 12 physics chapter 7 Alternating Current

NCERT solutions for class 12 physics chapter8 Electromagnetic Waves

NCERT solutions for class 12 physics chapter 9 Ray Optics and Optical Instruments

NCERT solutions for class 12 physics chapter 10 Wave Optics Solutions

NCERT solutions for class 12 physics chapter 11 Dual nature of radiation and matter

NCERT solutions for class 12 physics chapter 12 Atoms

NCERT solutions for class 12 physics chapter 13 Nuclei

NCERT solutions for class 12 physics chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits

Wave Optics Class 12 Solutions: Important Formulas and Diagrams

Huygens' Principle

Path difference (Δ):

T.D. (Time Difference):

Resultant Amplitude:

Resultant intensity:

Young’s Double Slit Experiment (YDSE)

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Path difference:

Where x be the position of point P from central maxima

For maxima at P:

For minima at P:

Brewster’s law:
Resolving power of Microscope:

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Resolving power of Telescope:

Also Check NCERT Books and NCERT Syllabus here:

Wave Optics NCERT Topics

The topics covered in the Wave Optics Class 12 NCERT Physics are Huygens Principle, Explanation of Refraction and Reflection using Huygens Principle, Addition of Waves, Interference, Defraction, Polarisation and related subtopics. One important experiment of Wave Optics is Young's Double Slit Experiment.

Importance of NCERT solutions for class 12 physics chapter 10 wave optics:

Previously, in the CBSE board exam, 11-mark questions were asked from the unit optics which includes chapters 9 Ray Optics and Wave Optics of NCERT Class 12 Physics.
For competitive exams also the Wave Optics NCERT solutions are important. Questions from the topic of young's double-slit experiment are frequently asked in competitive exams like NEET and JEE Main.
NCERT solutions for Class 12 Physics Chapter 10 wave optics are important to score well in exams.

Key features of Wave Optics Class 12 NCERT Solutions

Comprehensive Explanations: These wave optics class 12 exercise solutions offer comprehensive explanations, aiding students in understanding intricate wave optics concepts.
Problem-Solving Aid: Students can utilize these ncert class 12 physics chapter 10 pdf to practice and enhance their problem-solving skills in wave optics.
Clarity in Language: The solutions are presented in straightforward language, ensuring students can easily grasp and apply the principles.
Exam Preparation: These wave optics ncert solutions are a valuable resource for preparing for both board exams and competitive exams like JEE and NEET.
Performance Enhancement: By leveraging these solutions, students can improve their overall performance in physics.
Accessible for All: These wave optics class 12 solutions are freely accessible, making them available to all students seeking assistance with wave optics.

How to use NCERT solutions for class 12 Physics chapter 10 Wave Optics

Firstly go through the complete NCERT syllabus for Class 12 Physics and check all the topics. Through this, you will get a clear understanding of what to study.
Afterwards, try to solve the questions on your own before attempting the solutions.
If you still have any doubts then check the Wave Optics Class 12 NCERT Solutions.
Along with the Class 12 Physics Chapter 10 NCERT solutions, solve the previous year’s question papers and sample papers too.

NCERT solutions subject-wise

Frequently Asked Questions (FAQs)

1. Are NCERT books for Class 12 Physics enough to prepare for CBSE Board exams?

Yes, NCERT books are enough to prepare for the board exams, but you can refer to other reference books and sample papers as well. Try to cover all the concepts based on the NCERT syllabus. To get a good score in the CBSE board exam understand all the topics in the NCERT book and solve all the questions of NCERT exercise. Additionally students can refer NCERT exemplar problems and CBSE previous year question papers.

2. Will the NCERT solutions be helpful for competitive exams as well?

Yes, NCERT solutions will be helpful for competitive exams as well. Solving NCERT problems will give a better idea of concepts studied in a chapter and this in turn helps in competitive exams like JEE Main ans NEET.

3. Will the questions in the CBSE board exams be directly asked from the NCERT books?

Questions will be based on NCERT topics. But may be an application level question. The questions maynot be directly from the NCERT Questions but are related to the NCERT syllabus.

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NCERT Class 12 Maths Notes - Download Chapter wise Free PDFs

NCERT Class 12 Physics Chapter 9 Notes Ray Optics and Optical Instruments - Download PDF

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NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

Jefferson 25 September,2024

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

Read Complete Answer

Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Read Complete Answer

Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer

kumardurgesh1802 10 July,2024

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

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) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

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 .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

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.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

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

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

NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

NCERT Solutions for Class 12 Physics Chapter 10 – Download Free PDF

NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

NCERT solutions for Class 12 Pysics chapter 10 Wave Optics additional exercises:

Wave Optics Class 12 Solutions: Important Formulas and Diagrams

Huygens' Principle

Young’s Double Slit Experiment (YDSE)

Pearson | PTE

Also Check NCERT Books and NCERT Syllabus here:

Wave Optics NCERT Topics

Importance of NCERT solutions for class 12 physics chapter 10 wave optics:

Key features of Wave Optics Class 12 NCERT Solutions

How to use NCERT solutions for class 12 Physics chapter 10 Wave Optics

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