# NCERT Solutions for Class 12 Physics Chapter 10

## NCERT Solutions for Class 12 Physics Chapter 10 - Wave Optics NCERT Solutions for Wave Optics Class 12 are now available for students who want to get the best preparation for the exam. All the questions from this chapter along with the extra questions are solved with numerous examples. Experienced teachers prepare these essential NCERT Solutions. To obtain more marks in Physics, Download CBSE NCERT Solutions Class 12 Physics Wave Optics PDF. It is available in the free PDF format.

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1. Monochromatic light of wavelength $589nm$is incident from air on a water surface. What are the wavelength, frequency and speed of

(a) Reflected light?

Ans: Wavelength of incident monochromatic light is given as,

$\lambda =589nm=589\times {{10}^{9}}m$

Speed of light in air is $c=3\times {{10}^{8}}m$

Refractive index of water is $\mu =1.33$

In this case, the ray will reflect in the same medium as that of the incident ray. Hence, the wavelength, speed, and frequency of the reflected ray will be the equal to that of the incident ray.

Frequency of light can be given by the relation,

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

Where,

$\nu =$Frequency of light

$c=$Speed of light

$\lambda =$Wavelength of light

$\Rightarrow \nu =\frac{3\times {{10}^{8}}}{589\times {{10}^{-9}}}$

$\Rightarrow \nu =5.09\times {{10}^{14}}Hz$

Hence, the speed, frequency, and wavelength of the reflected light are

$3\times {{10}^{8}}m/s$ , $5.09\times {{10}^{14}}Hz$ and $589nm$respectively.

(b) Refracted light? Refractive index of water is $1.33$.

Ans: The frequency of light is independent of the property of the medium in which it is travelling.

Hence, the frequency of the refracted ray in water will be equal to the frequency of the incident light or reflected light in air.

Frequency of the refracted light ray, $\nu =5.09\times {{10}^{14}}Hz$

Speed of light in water is related to the refractive index of water as given in the formula below:

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

$\Rightarrow v=\frac{3\times {{10}^{8}}}{1.33}$

$\Rightarrow v=2.26\times {{10}^{8}}m/s$

The formula below gives the relation of wavelength of light in water and the speed and frequency of light,

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

$\Rightarrow \lambda =\frac{2.26\times {{10}^{8}}}{5.09\times {{10}^{14}}}$

$\Rightarrow \lambda =444.007\times {{10}^{-9}}m$

$\Rightarrow \lambda =444.01nm$

2. What is the shape of the wavefront in each of the following cases:

(a) Light diverging from a point source.

Ans: When a light diverges from a point source, the shape of the wavefront in this case is spherical. The wavefront originating from a point source is shown in the given figure.

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

Ans: The shape of the wavefront when a light emerges out of a convex lens when a point source is placed at its focus is a parallel grid. This can be represented as shown in the given figure.

(c) The portion of the wavefront of light from a distant star intercepted by the Earth.

Ans: In this case the portion of the wavefront of light from a distant star intercepted by the Earth is a plane.

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

Ans: Refractive index of glass is given as,

$\mu =1.5$

Speed of light, $c=3.0\times {{10}^{8}}m/s$

Speed of light in glass is given by the formula,

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

$\Rightarrow v=\frac{3\times {{10}^{8}}}{1.5}=2\times {{10}^{8}}m/s$

Hence, the speed of light in glass is $2\times {{10}^{8}}m/s$.

(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?

Ans: The speed of light in glass depends on the colour of light.

The refractive index of a violet component of white light is more than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass as speed and refractive index are inversely related to each other.

Hence, violet light travels slower as compared to red light in a glass prism.

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

Ans: Distance between the slits is given as, $d=0.28mm=0.28\times {{10}^{-3}}m$

Distance between the slits and the screen, $D=1.4m$

Distance between the central fringe and the fourth $\left( n=4 \right)$fringe, $u=1.2cm=1.2\times {{10}^{-2}}m$

In case of a constructive interference, the relation for the distance between the two fringes can be given as: $u=n\lambda \frac{D}{d}$

where,

$n=$ Order of fringes $=4$

$\lambda =$ Wavelength of light used

$\lambda =\frac{ud}{nD}$

$\Rightarrow \lambda =\frac{1.2\times {{10}^{-2}}\times 0.28\times {{10}^{-3}}}{4\times 1.4}$

$\Rightarrow \lambda =6\times {{10}^{-7}}$

$\Rightarrow \lambda =600nm$

Hence, the wavelength of the light is $600nm$.

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 $\frac{\lambda }{3}$?

Ans: The intensity of the two light waves be ${{I}_{1}}$ and ${{I}_{2}}$. Their resultant intensities can be evaluated as: ${I}'={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}\cos \phi$

Where,

$\phi =$The phase difference between two waves for monochromatic light waves,

Since ${{I}_{1}}={{I}_{2}}$

So ${I}'=2{{I}_{1}}+2{{I}_{1}}\cos \phi$

The formula for phase difference can be given as:

$Phase\,difference=\frac{2\pi }{\lambda }\times \,Path\,difference$

Since, path difference is $\lambda$,

Phase difference is $\phi =2\pi$

${I}'=2{{I}_{1}}+2{{I}_{1}}=4{{I}_{1}}$

Given,

${{I}_{1}}=\frac{{{K}'}}{4}$ …… (1)

When path difference $=\frac{\lambda }{3}$

phase difference, $\phi =\frac{2\pi }{3}$

Hence, resultant intensity,

${{I}_{R}}={{I}_{1}}+{{I}_{1}}+2\sqrt{{{I}_{1}}{{I}_{1}}}\cos \frac{2\pi }{3}$

$\Rightarrow {{I}_{R}}=2{{I}_{1}}+2{{I}_{2}}\left( -\frac{1}{2} \right)={{I}_{1}}$

Using equation (1), we can state that

${{I}_{R}}={{I}_{1}}=\frac{K}{4}$

Hence, for monochromatic light waves, the intensity of light at a point where the path difference is $\frac{\lambda }{3}$ is $\frac{K}{4}$units.

6. A beam of light consisting of two wavelengths, $650nm$ and $520nm$, 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 $650nm$.

Ans: Given that,

Wavelength of the first light beam, ${{\lambda }_{1}}=650nm$

Wavelength of second light beam, ${{\lambda }_{2}}=520nm$

Distance of the slits from the screen $=D$

Distance between the two slits $=d$

Distance of the ${{n}^{th}}$ bright fringe on the screen from the central maximum is given by the formula below,

$x=n{{\lambda }_{1}}\left( \frac{D}{d} \right)$

For the third bright fringe, $n=3$

$x=3\times 650\times \frac{D}{d}=1950\left( \frac{D}{d} \right)nm$, which is nothing but the distance of the third bright fringe on the screen from the central maximum.

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

Ans: In this case, let the ${{n}^{th}}$bright fringe due to wavelength ${{\lambda }_{2}}$and ${{\left( n-1 \right)}^{th}}$ bright fringe due to wavelength ${{\lambda }_{1}}$ coincide on the screen. Equate the conditions for bright fringes as follows:

$n{{\lambda }_{2}}=\left( n-1 \right){{\lambda }_{1}}$

$\Rightarrow 520n=650n-650$

$\Rightarrow 650=130n$

$\Rightarrow n=5$

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

$x=n{{\lambda }_{2}}\frac{D}{d}$

$\Rightarrow x=5\times 520\frac{D}{d}=2600\frac{D}{d}nm$

Note: The value of $d$ and $D$ are not given in the question, hence the exact answer cannot be found.

7. In a double-slit experiment the angular width of a fringe is found to be

$0.2{}^\circ$ on a screen placed $1m$ away. The wavelength of light used is $600nm$. 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 $\frac{4}{3}$.

Ans: Given that,

Distance of the screen from the slits is given as, $D=1m$

Wavelength of light used, ${{\lambda }_{1}}=600nm$

Angular width of the fringe in air, ${{\theta }_{1}}=0.2{}^\circ$

Angular width of the fringe in water  $={{\theta }_{2}}$

Refractive index of water is $\frac{4}{3}$.

Refractive index is associated with angular width as:

$\mu =\frac{{{\theta }_{1}}}{{{\theta }_{2}}}$

$\Rightarrow {{\theta }_{2}}=\frac{3}{4}{{\theta }_{1}}$

$\Rightarrow {{\theta }_{2}}=\frac{3}{4}\times 0.2=0.15{}^\circ$

Therefore, the angular width of the fringe in this case $\theta$ in water will reduce to $0.15{}^\circ$.

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

Ans: Given that,

Refractive index of glass is given $=1.5$

Brewster angle $=\theta$

Brewster angle is associated to refractive index as mentioned in the formula below:

$\tan \theta =\mu$

$\Rightarrow \theta ={{\tan }^{-1}}\left( 1.5 \right)=56.31{}^\circ$

Hence, the Brewster angle for transition from air to glass is $56.31{}^\circ$.

9. Light of wavelength $5000{{A}^{{}^\circ }}$ 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?

Ans: Wavelength of incident light is given,

$\lambda =5000\overset{{}^\circ }{\mathop{A}}\,=5000\times {{10}^{-10}}m$

Speed of light,

$c=3\times {{10}^{8}}m/s$

Frequency of incident light is given by the formula,

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

$\Rightarrow \nu =\frac{3\times {{10}^{8}}}{5000\times {{10}^{10}}}=6\times {{10}^{14}}Hz$

The wavelength and frequency of incident light is equal to that of reflected ray.

Therefore, the wavelength of reflected light is $5000{{A}^{{}^\circ }}$and its frequency is $6\times {{10}^{14}}Hz$.

When reflected ray is normal to incident ray, the addition of the angle of incidence, given as, $\angle i$ and angle of reflection, given as, $\angle r$ is ${{90}^{\circ }}$.

According to the law of reflection, the angle of incidence is always the same as the angle of reflection. Hence, the sum can be written as:

$\angle i+\angle r={{90}^{\circ }}$

$\Rightarrow \angle i+\angle i={{90}^{\circ }}$

$\Rightarrow 2\angle i={{90}^{\circ }}$

$\Rightarrow \angle i=\frac{{{90}^{\circ }}}{2}={{45}^{\circ }}$

Hence, the angle of incidence for the given condition in the question is ${{45}^{\circ }}$.

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

Ans: Fresnel’s distance $\left( {{Z}_{F}} \right)$ can be defined as the distance for which the ray optics is a good approximation. It can be expressed by the relation,

${{Z}_{F}}=\frac{{{a}^{2}}}{\lambda }$

Where,

Aperture width is $a$,

Wavelength of light is $\lambda$.

Now, $a=4mm=4\times {{10}^{-3}}m$

$\lambda =400nm=400\times {{10}^{-9}}m$

On substitution,

${{Z}_{F}}=\frac{{{\left( 4\times {{10}^{-3}} \right)}^{2}}}{400\times {{10}^{-9}}}=40m$

Thus, the distance for which the ray optics is a good approximation is $40m$.

11. The $6563\overset{{}^\circ }{\mathop{A}}\,\,{{H}_{a}}$ line emitted by hydrogen in a star is found to be red shifted by $15\overset{{}^\circ }{\mathop{A}}\,$. Estimate the speed with which the star is receding from the Earth.

Ans: Wavelength of ${{H}_{a}}$line emitted by hydrogen is given as,

$\lambda =6563\overset{{}^\circ }{\mathop{A}}\,$

$\Rightarrow \lambda =6563\times -{{10}^{-10}}m$

Red shift of the star can be written as,

$\left( {\lambda }'-\lambda \right)=15\overset{{}^\circ }{\mathop{A}}\,=15\times {{10}^{-10}}m$

Speed of light,

$c=3\times {{10}^{8}}m/s$

Let us suppose that the velocity of the star receding away from the Earth be $v$.

The red shift is associated with velocity as given:

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

$\Rightarrow v=\frac{c}{\lambda }\times \left( {\lambda }'-\lambda \right)$

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

Therefore, the speed with which the star is receding away from the Earth is $6.87\times {{10}^{5}}m/s$.

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?

Ans:  No, the prediction is not confirmed by the experimental determination of speed of light in water and the theory which is an alternative picture of light is consistent with wave theory.

Corpuscular theory of light given by Newton suggests that when light corpuscles strike the interface of two media from a rarer (air) to a denser (water) medium, forces of attraction normal to the surface are experienced by the particles. Thus, the normal component of velocity increases while the component along the surface does not change.

Clearly, the expression can be written as:

$c\sin i=v\sin r$ …… (1)

Where,

$i=Angle\,of\,incidence$

$r=Angle\,of\,reflection$

$c\,\,=Velocity\,of\,light\,in\,air$

$v=$Velocity of light in water

We have the expression for relative refractive index of water with respect to air as:

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

Hence, equation (1) reduces to

$\Rightarrow \frac{v}{c}=\frac{\sin i}{\sin r}=\mu$ …… (2)

But $\mu >1$

Hence, it can be observed from equation (2) that $v>c$.

This is not possible because this prediction is opposite to the experimental results of $c>v$ .

Clearly, the wave picture of light is consistent with the experimental results.

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.

Ans: Let an object at $O$be placed in front of a plane mirror $M{O}'$at a distance $r$ (as shown in the adjoining figure).

A circle is drawn from the centre ($O$) such that it just touches the plane mirror at point ${O}'$on the mirror.

According to Huygens' Principle, $XY$ is the wavefront of incident light.

If the mirror is not present, then a similar wavefront ${X}'{Y}'$ (as $XY$) would form behind the point ${O}'$ at distance $r$ (as shown in the given figure above).

${X}'{Y}'$ can be considered as a virtual reflected ray for the plane mirror.

Therefore, a point object placed in front of the plane mirror generates a virtual image whose distance from the mirror is the same as the object distance ($r$).

14. 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,

Ans: The speed of light in a vacuum i.e., $c=3\times {{10}^{8}}m/s$ (approximately) is a universal constant.

The motion of the source, the observer, or both, does not affect the speed of light. Hence, the above given factors do not affect the speed of light in a vacuum.

(b) The speed of light in a medium (say, glass or water), depends?

Ans:  Among the above listed factors, the speed of light in a medium is dependent on the wavelength of light in that medium.

15. For sound waves, the Doppler formula for frequency shift differs slightly between the two situations:

(a) source at rest; observer moving, and

(b) 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?

Ans: No, the formulae for the two situations cannot be strictly identical.

Sound waves need a medium for propagation. The two given circumstances are not identical scientifically since the motion of an observer relative to a medium is not similar in the two situations. Hence, the doppler formulas for the two situations are different.

In case of light waves, they can travel in vacuum. In vacuum, the above two situations are identical because the speed of light is not dependent of the motion of the observer and the motion of the source when light travels in a medium. The above two cases are not identical because the speed of light depends on the wavelength of the medium.

16. In double-slit experiment using light of wavelength $600nm$, the angular width of a fringe formed on a distant screen is $0.1{}^\circ$. What is the spacing between the two slits?

Ans: Wavelength of light used is given as,

$\lambda =600nm=600\times {{10}^{-9}}m$

Angular width of fringe is,

$\theta =0.1{}^\circ =0.1\times \frac{\pi }{180}=\frac{3.14}{1800}rad$

Angular width of a fringe is related to slit spacing ($d$) as:

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

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

$\Rightarrow d=\frac{600\times {{10}^{-9}}}{\frac{3.14}{1800}}=3.44\times {{10}^{-4}}m$

Hence, the spacing between the slits is $3.44\times {{10}^{-4}}m$.

(a) 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?

Ans: In a single slit diffraction experiment, if the width of the slit is made two times the original width, then the size of the central diffraction band reduces to half times and the intensity of the central diffraction band increases up to four times.

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

Ans: The interference pattern obtained in a double-slit experiment is modulated by diffraction from each slit. The pattern is the outcome of the interference of the diffracted wave from each slit.

(c) 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?

Ans: When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is observed at the centre of the shadow of the obstacle. This occurs because the light waves are diffracted from the edge of the circular obstacle, which is interfered constructively at the centre of the shadow. This constructive interference makes a bright spot.

(d) Two students are separated by a $7m$ partition wall in a room $10m$ 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.

Ans: Waves can be bent by obstacles by a large angle, which becomes possible when the size of the obstacle is comparable to the wavelength of the waves.

On the contrary, the wavelength of the light waves is too small in comparison to the size of the obstacle. Hence, the angle of diffraction will be less. Hence, the students are not able to see each other. On the contrary, the size of the wall is comparable to the wavelength of the sound waves. Thus, the bending of the waves takes place at a great angle. Therefore, the students are able to hear each other.

(e) 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?

Ans: The justification for the above statement is that in ordinary optical instruments, the size of the aperture involved is much bigger than the wavelength of the light used.

18. Two towers on top of two hills are $40km$ apart. The line joining them passes $50m$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?

Ans: Given that,

Distance between the towers is given as $d=40km$

Height of the line joining the hills is $50m$

Therefore, the radial spread of the radio waves should not exceed $50km$.

Since the hill is located halfway between the towers, Fresnel’s distance can be written as: ${{Z}_{p}}=20km=20\times {{10}^{4}}m$

Aperture can be written as:

$a=d=50m$

Fresnel’s distance is given by the expression,

${{Z}_{p}}=\frac{{{a}^{2}}}{\lambda }$

Where,

$\lambda =$ Wavelength of radio waves

$\lambda =\frac{{{a}^{2}}}{{{Z}_{p}}}$

$\Rightarrow \lambda =\frac{{{\left( 50 \right)}^{2}}}{2\times {{10}^{4}}}=1250\times {{10}^{-4}}=0.1250m$

$\Rightarrow \lambda =12.5cm$

Therefore, the wavelength of the radio waves is $12.5cm$.

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

Ans: Given that,

Wavelength of light beam is $\lambda =500nm=500\times {{10}^{-9}}m$

Distance of the screen from the slit, $D=1m$

For first minima, $n=1$

Distance between the slits $=d$

Distance of the first minimum from the centre of the screen can be given as:

$x=2.5mm=2.5\times {{10}^{-3}}m$

It is related to the order of minima as given in the formula:

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

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

$\Rightarrow d=\frac{1\times 500\times {{10}^{-9}}\times 1}{2.5\times {{10}^{-3}}}$

$\Rightarrow d=2\times {{10}^{-4}}m=0.2mm$

Clearly, the width of the slits is $0.2mm$.

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

Ans: Weak radar signals which are sent by a low flying aircraft can interfere with the TV signals received by the TV antenna. Due to this, the TV signals may get distorted. Hence, when a low flying aircraft passes overhead, we sometimes observe a slight shaking of the picture on our TV screen.

(b) 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?

Ans: The principle of linear superposition of wave displacement is required to our understanding of intensity distributions and interference patterns. This holds true because superposition follows from the linear character of a differential equation that governs wave motion. If ${{y}_{1}}$ and ${{y}_{2}}$are the solutions of the second order wave equation, then a linear combination of ${{y}_{1}}$ and ${{y}_{2}}$  and thus, will also be the solution of the wave equation.

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

Ans: Let a single slit of width $d$is divided into $n$ smaller slits.

Width of each slit can be expressed as ${d}'=\frac{d}{n}$

Angle of diffraction is given by the formula,

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

Now, each of these infinitesimally small slits transmit zero intensity in direction $\theta$. Therefore, the combination of these slits would give zero intensity.

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### Class 12: Wave Optics

They cover all exercises given below:

### NCERT Solutions for Class 12 Physics Chapter 10 – Wave Optics

 Ex - 1 Introduction Ex - 2 Refraction And Reflection of Plane Waves Using Huygens Principle Ex - 3 Refraction of a Plane Wave Ex - 4 Refraction At a rarer medium Ex - 5 Reflection of a Plane Wave By a Plane Surface Ex - 6 The Doppler Effect, And Coherent And Incoherent Addition of Waves Ex - 7 Interference of Light Waves And Young’s Experiment Ex - 8 Diffraction Ex - 9 The Single Slit, Seeing The Single Slit Diffraction Pattern Ex - 10 Resolving Power of Optical Instruments Ex - 11 The Validity of Ray Optics Ex - 12 Polarisation, Polarisation By Scattering and Polarisation By Reflection

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Q1. What are Optical Waves?

Ans. In Physics, Wave Optics generally refers to the branch of the optic from which interference, polarization, diffraction is studied. Magnetic fields and electrical fields of high frequency are combined to be termed as Optical Waves. The wave ranges from the infrared to the ultraviolet light. It helps to analyze the propagation of the light full of photons.

Q2. What are the Two Types of Optics?

Ans. Classical optics are mainly classified into two main parts. The first one is Ray or Geometrical optic and another one is physical or waves optics. The theory in which light is considered to travel a straight line comes under geometrical optics. The physical or Wave Optics include the light considered as an Electromagnetic way. As a whole in Physics, light normally indicates electromagnetic radiation of any wavelength, whether it is visible or not.

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