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Chapter 10: Wave Optics

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, \(\lambda\) = 589 nm = 589 x 10-9 m

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

Refractive index of water, \(\mu\) = 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:

\(v=\frac{c}{\lambda}\) = \(\frac{3\times10^{8}}{589\times10^{-9}}\) = 5.09 × 1014 Hz

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

c = 3 x 108 m s-1, 5.09 × 1014 Hz, and 589 nm respectively.

 

(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

Speed of light in water Is related to the refractive Index of water as:

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

Wavelength of light in water can be found by the relation:

\(\lambda=\frac{v}{V}\) = \(\frac{2.26\times10^{8}}{5.09\times10^{14}}\) = 444.007 × 10-9 m = 444.01nm

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

 

Question 2:

Find the shape of the wave front 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 wave front of light from a distant star intercepted by the Earth.

Answer:

(i) The shape of a wave front is spherical in the case of a light diverging from a point source. The wave front is shown in the figure

 

Wave Optics

 

(ii) The shape of a wave front is a parallel odd in the case of a light emerging out of a convex lens when a point source is placed at its focus.

 

Wave front

 

(iii) The shape of the wave front is a plane when a portion of the wave front of light from a distant star intercepted by the earth.

 

Question 3:

(i) The refractive index of glass is 1.5. What is the speed of light in glass? Speed of light in 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, \(\mu\) = 1.5

Speed of light, c = 3 × 108 ms-1

Speed of light in glass is given by the relation; \(v=\frac{c}{\mu}\)

= \(\frac{3\times10^{8}}{1.5}\) = \(2\times 10^{8}m/s\)

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

(ii) The speed of light in glass is not independent of the colour of light. The refractive Index of a violet component of white light is greater than the refractive Index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.

 

 

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

The relation for the distance between the two fringes as in case of a constructive interference:

\(u=n\;\lambda\;\frac{D}{d}\)

Where, n = order of fringes

= 4\(\lambda\) = Wavelength of light used

\(u=n\;\lambda\;\frac{D}{d}\)

= \(\frac{1.2\times10^{-2}\times 0.28\times 10^{-3}}{4\times 1.4}\)

= 60 × 10-7 = 600nm

Therefore the wavelength of the light is 600 nm.

 

Question 5:

In Young’s double-slit experiment using the 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:

Let \(I_{1}\) and \(I_{2}\) be the intensity of the two light waves. Their resultant intensities can be obtained as:

\(I’=I_{1}+I_{2}+2\sqrt{I_{1}\;I_{2}}\; cos\phi\)

Where,

\(\phi\) = Phase difference between the two waves

For monochromatic light waves:

\(I_{1}\) = \(I_{2}\)

Therefore \(I’=I_{1}+I_{2}+2\sqrt{I_{1}\;I_{2}}\; cos\phi\)

= \(2I_{1}+2I_{1}\;cos\phi\)

Phase difference = \(\frac{2\pi}{\lambda}\times\;Path\;difference\)

Since path difference = \(\lambda\), Phase difference, \(\phi=2\pi\) and I’ = K [Given]

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

When path difference= \(\frac{\lambda}{3}\)

Phase difference, \(\phi=\frac{2\pi}{3}\)

Hence, resultant intensity:

\(I’_{g}=I_{1}+I_{1}+2\sqrt{I_{1}\;I_{1}}\; cos\frac{2\pi}{3}\)

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

Using equation (i), we can write:

\(I_{g}=I_{1}=\frac{K}{4}\)

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

 

Question 6: 650 nm and 520 nm are two wavelengths of a beam of light which is used to obtain interference fringes in 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, \(\lambda_{1}\) = 650 nm

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

Distance of the slits from the screen = D

Distance between the two slits = d

(i) Distance of the \(n^{th}\) bright fringe on the screen from the central maximum is given by the relation,

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

For third bright fringe, n=3

Therefore x = \(3\times650\frac{D}{d}= 1950\frac{D}{d}nm\)

(b) Let, the \(n^{th}\) bright fringe due to wavelength \(\lambda_{2}\) and \((n – 1)^{th}\) bright fringe due to wavelength \(\lambda_{2}\) 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

Therefore n = 5

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

x = \(n\;\lambda_{2}\;\frac{D}{d}\) = \(5\times 520\frac{D}{d}=2600\frac{D}{d}\) 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 \(\frac{4}{3}\).

Answer:

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

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

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

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

Refractive index of water, \(\mu=\frac{4}{3}\)

Refractive index is related to angular width as:

\(\mu=\frac{\theta_{1}}{\theta_{2}}\) \(\theta_{2}=\frac{3}{4}\theta_{1}\) \(\frac{3}{4}\times 0.2=0.15\)

Therefore, the angular width of the fringe in water will reduce to 0.15°

 

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

 Answer:

Refractive index of glass, \(\mu=1.5\)

Let, Brewster angle = \(\theta\)

Brewster angle is related to refractive index as:

\(tan\theta=\mu\)

\(\theta=tan^{-1}(1.5)\) = 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, \([\lambda]\) = 5000 Armstrong = 5000 x 10-10 m

Speed of light, c =3 x 108 m

Frequency of incident light is given by the relation,

v = \(\frac{c}{\lambda}\) = \(\frac{3\;\times \;10^{8}}{5000\;\times \;10^{-10}}\) = 6 × 1014

The wavelength and frequency of incident light is the same as that of reflected ray. Hence, the wavelength of reflected light is 5000 Armstrong and its frequency is \(6 \times 10^{14}\)Hz. When reflected ray is normal to incident ray, the sum of the angle of incidence, \(\angle i\) and angle of reflection, \(\angle r\) is 90°

According to the law of reflection, the angle of incidence is always equal to the angle of reflection. Hence, we can write the sum as:

\(\angle i+\angle r\) = 90°

i.e. \(\angle i+\angle i\) = 90°

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

Therefore, the angle of incidence for the given condition is 45°

 

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 (\(Z_{F}\)) is the distance for which the ray optics is a good approximation. It is given by the relation,

\(Z_{F}=\frac{a^{2}}{\lambda}\)

Where,

Aperture width, a = 4 mm = 4 × 10-3 m

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

\(Z_{F}=\frac{(4\times10^{-3})^{2}}{400\times10^{-9}}\) = 40m

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

 



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