NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics is an essential study material that every student requires for effective preparation of Class 12 board examination and graduation entrance examination. Wave optics Class 12 physics NCERT solutions pdf provides answers to the question in textbooks, previous year question papers, and sample papers.

NCERT solutions Class 12 Physics Chapter 10 Wave Optics PDF comprises of MCQs, exemplary problems, worksheets and exercises that help you understand the topic clearly, ultimately, helping you score good marks in Class 12 as well as entrance examination.

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Class 12 Physics NCERT Solutions for Chapter 10 Wave Optics

The derivation of laws of refraction and reflection using Huygens law is often asked in exams. The Brewster law and the angular width expression of the central maximum of the diffraction pattern are frequently asked. Besides the derivation and the theory, the numerical problems from various topics are asked in the exams. All the topics in the chapter are covered in the NCERT Solutions provided here. Download the free PDF provided here, if necessary, take a printout to keep it handy during the preparation of exams.

Topics covered in Class 12 Physics Chapter 10 Wave Optics

Section Number Topic
10.1 Introduction
10.2 Huygens Principle
10.3 Refraction And Reflection Of Plane Waves Using Huygens Principle
10.3.1 Refraction Of A Plane Wave
10.3.2 Refraction At A Rarer Medium
10.3.3 Reflection Of A Plane Wave By A Plane Surface
10.3.4 The Doppler Effect
10.4 Coherent And Incoherent Addition Of Waves
10.5 Interference Of Light Waves And Young’s Experiment
10.6 Diffraction
10.6.1 The Single Slit
10.6.2 Seeing The Single Slit Diffraction Pattern
10.6.3 Resolving Power Of Optical Instruments
10.6.4 The Validity Of Ray Optics
10.7 Polarisation
10.7.1 Polarisation By Scattering
10.7.2 Polarisation By Reflection

 

Class 12 Physics NCERT Solutions Wave Optics Important Questions


Question 1:

Monochromatic light having a wavelength of 589nm from the air is incident on a water surface. Find the frequency, wavelength and speed of (i) reflected and (ii) refracted light? [1.33 is the Refractive index of water]

Answer:

Monochromatic light incident having wavelength, λ\lambda = 589 nm = 589 x 10-9 m

Speed of light in air, c = 3 x 108 m s-1

Refractive index of water, μ\mu = 1.33

(i) In the same medium through which incident ray passed the ray will be reflected back.

Therefore the wavelength, speed, and frequency of the reflected ray will be the same as that of the incident ray.

Frequency of light can be found from the relation:

v=cλv=\frac{c}{\lambda} = 3×108589×109\frac{3\times10^{8}}{589\times10^{-9}} = 5.09 × 1014 Hz

Hence, c = 3 x 108 m s-1, 5.09 × 1014 Hz, and 589 nm are the speed, frequency, and wavelength of the reflected light.

(b) The frequency of light which is travelling never depends upon the property of the medium. Therefore, the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected light in air.

Refracted frequency, v = 5.09 x 1014 Hz

Following is the relation between the speed of light in water and the refractive index of the water:

v=cλv=\frac{c}{\lambda} = v=3×1081.33v=\frac{3\times10^{8}}{1.33} = 2.26 × 108 m s‑1

Below is the relation for finding the wavelength of light in water:

λ=vV\lambda=\frac{v}{V} = 2.26×1085.09×1014\frac{2.26\times10^{8}}{5.09\times10^{14}} = 444.007 × 10-9 m = 444.01nm

Therefore, 444.007 × 10-9 m, 444.01nm, and 5.09 × 1014 Hz  are the speed, frequency, and the wavelength of the refracted light.

 Question 2:

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

(i) Light diverging from a point source.

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

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

Answer:

(i) When the light diverges from a point source, the shape of the wavefront is spherical. Following is the figure of the wavefront:

 

Wave Optics

 

(ii) When the light is emerging from the convex lens, the shape of the wavefront is parallel odd. In this case, the point source is placed at its focus.

 

Wave front

 

(iii) When the light is coming from a distant star that is intercepted by the earth, the shape of the wavefront is plane.

Question 3:

(i) The refractive index of glass is 1.5. What is the speed of light in glass? Speed of light in a vacuum is ( 3.0 x 108 m s-1 )

(ii) 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:

(i) Refractive Index of glass, μ\mu = 1.5

Speed of light, c = 3 × 108 ms-1

The relation for the speed of light in a glass is: v=cμv=\frac{c}{\mu}

= 3×1081.5\frac{3\times10^{8}}{1.5} = 2×108m/s2\times 10^{8}m/s

Hence, the speed of light in glass is 2 × 108 m s-1

(ii) The speed of light is dependent on the colour of the light. For a white light, the refractive index of the violet component is greater than the refractive index of the red component. So, the speed of violet light is less than the speed of the red light in the glass. This will reduce the speed of violet light in glass prism when compared with red light.

Question 4:

In Young’s double-slit experiment, 0.28mm separation between the slits and the screen is placed 1.4m away. 1.2cm is the distance between the central bright fringe and the fourth bright fringe. Determine the wavelength of light used in the experiment.

Answer:

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

and the distance between the slits, d = 0.28 mm = 0.28 x 10-3 m

Distance between the central fringe and the fourth (n = 4) fringe,

u = 1.2cm = 1.2 × 10-2 m

For a constructive interference, following is the relation for distance between the two fringes:

u=n  λ  Ddu=n\;\lambda\;\frac{D}{d}

Where, n = order of fringes

= 4λ\lambda = Wavelength of light used

u=n  λ  Ddu=n\;\lambda\;\frac{D}{d}

= 1.2×102×0.28×1034×1.4\frac{1.2\times10^{-2}\times 0.28\times 10^{-3}}{4\times 1.4}

= 60 × 10-7 = 600nm

600nm is the wavelength of the light.

Question 5:

In Young’s double-slit experiment using the monochromatic light of wavelength λ\lambda, the intensity of light at a point on the screen where path difference is λ\lambda, is K units. What is the intensity of light at a point where path difference is λ3\frac{\lambda}{3}?

Answer:

Let I1I_{1} and I2I_{2} be the intensity of the two light waves. Their resultant intensities can be obtained as:

I=I1+I2+2I1  I2  cosϕI’=I_{1}+I_{2}+2\sqrt{I_{1}\;I_{2}}\; cos\phi

Where,

ϕ\phi = Phase difference between the two waves

For monochromatic light waves:

I1I_{1} = I2I_{2}

Therefore I=I1+I2+2I1  I2  cosϕI’=I_{1}+I_{2}+2\sqrt{I_{1}\;I_{2}}\; cos\phi

= 2I1+2I1  cosϕ2I_{1}+2I_{1}\;cos\phi

Phase difference = 2πλ×  Path  difference\frac{2\pi}{\lambda}\times\;Path\;difference

Since path difference = λ\lambda, Phase difference, ϕ=2π\phi=2\pi and I’ = K [Given]

Therefore I1=K4I_{1}=\frac{K}{4} . . . . . . . . . . . . . . . (i)

When path difference= λ3\frac{\lambda}{3}

Phase difference, ϕ=2π3\phi=\frac{2\pi}{3}

Hence, resultant intensity:

Ig=I1+I1+2I1  I1  cos2π3I’_{g}=I_{1}+I_{1}+2\sqrt{I_{1}\;I_{1}}\; cos\frac{2\pi}{3}

= 2I1+2I1(12)\\2I_{1}+2I_{1}(-\frac{1}{2})

Using equation (i), we can write:

Ig=I1=K4I_{g}=I_{1}=\frac{K}{4}

Hence, the intensity of light at a point where the path difference is λ3\frac{\lambda}{3} is K4\frac{K}{4} units.

Question 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.

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

 Answer:

Wavelength of the light beam, λ1\lambda_{1} = 650 nm

Wavelength of another light beam, λ2\lambda_{2} = 520 nm

Distance of the slits from the screen = D

Distance between the two slits = d

(i) Distance of the nth bright fringe on the screen from the central maximum is given by the relation,

x =n  λ1  (Dd)n\;\lambda_{1}\;(\frac{D}{d})

For third bright fringe, n=3

Therefore x = 3×650Dd=1950Ddnm3\times650\frac{D}{d}= 1950\frac{D}{d}nm

(b) Let, the nthn^{th} bright fringe due to wavelength λ2\lambda_{2} and (n1)th(n – 1)^{th} bright fringe due to wavelength λ2\lambda_{2} coincide on the screen. The value of n can be obtained by equating the conditions for bright fringes:

nλ2=(n1)λ1n\lambda_{2}=(n-1)\lambda_{1}

520n = 650n – 650

650=130n

Therefore n = 5

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

x = n  λ2  Ddn\;\lambda_{2}\;\frac{D}{d} = 5×520Dd=2600Dd5\times 520\frac{D}{d}=2600\frac{D}{d} nm

Note: The value of d and D are not given in the question.

Question 7:

In a double-slit experiment, 0.2° is found to be the angular width of a fringe 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 43\frac{4}{3}.

Answer:

Distance of the screen from the slits, D=1m

Wavelength of light used, λ1\lambda_{1} = 600 nm

Angular width of the fringe in air θ1\theta_{1} = 0.2°

Angular width of the fringe in water=θ2\theta_{2}

Refractive index of water, μ=43\mu=\frac{4}{3} μ=θ1θ2\mu=\frac{\theta_{1}}{\theta_{2}} is the relation between the refractive index and the angular width

θ2=34θ1\theta_{2}=\frac{3}{4}\theta_{1} 34×0.2=0.15\frac{3}{4}\times 0.2=0.15

Therefore, 0.15° is the reduction in the angular width of the fringe in water.

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

 Answer:

Refractive index of glass, μ=1.5\mu=1.5

Consider Brewster angle = θ\theta

Following is the relation between the Brewster angle and the refractive index:

tanθ=μtan\theta=\mu θ=tan1(1.5)\theta=tan^{-1}(1.5) = 56.31°

Therefore, the Brewster angle for air to glass transition is 56.31°

Question 9: Light of wavelength 5000 Armstrong 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:

Wavelength of incident light, [λ][\lambda] = 5000 Armstrong = 5000 x 10-10 m

Speed of light, c =3 x 108 m

Following is the relation for the frequency of incident light:

v = cλ\frac{c}{\lambda} = 3  ×  1085000  ×  1010\frac{3\;\times \;10^{8}}{5000\;\times \;10^{-10}} = 6 × 1014

The wavelength and frequency of incident light is equal to the reflected ray. Therefore, 5000 Armstrong and 6×10146 \times 10^{14}Hz is the wavelength and frequency of the reflected light. When reflected ray is normal to incident ray, the sum of the angle of incidence, i\angle i and angle of reflection, r\angle r is 90°.

From laws of reflection we know that the angle of incidence is always equal to the angle of reflection

i+r\angle i+\angle r = 90°

i.e. i+i\angle i+\angle i = 90°

Hence, i=902\angle i=\frac{90}{2} = 45°

Therefore, 45° is the angle of incidence.

Question 10:

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

 Answer:

Fresnel’s distance (ZFZ_{F}) is the distance which is used in ray optics for a good approximation. Following is the relation,

ZF=a2λZ_{F}=\frac{a^{2}}{\lambda}

Where,

Aperture width, a = 4 mm = 4 × 10-3 m

Wavelength of light, λ\lambda = 400 nm = 400 × 10-9 m

ZF=(4×103)2400×109Z_{F}=\frac{(4\times10^{-3})^{2}}{400\times10^{-9}} = 40m

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

 

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