NCERT Solutions Class 12 Physics Chapter 10: Wave Optics

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

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

Refractive index of water, μ = 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:

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:

Wavelength of light in water is given by the relation,

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

 

(a) The shape of the wavefront in case of a light diverging from a point source is spherical. The wavefront emanating from a point source is shown in the given figure.

(b) The shape of the wavefront in case of a light emerging out of a convex lens when a point source is placed at its focus is a parallel grid. This is shown in the given figure.

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

 

(a) Refractive index of glass, μ = 1.5

Speed of light, c = 3 × 108 m/s

Speed of light in glass is given by the relation,

Hence, the speed of light in glass is 2 × 108 m/s.

(b) 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.

 

Distance between the slits, d = 0.28 mm = 0.28 × 10−3 m

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

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

u = 1.2 cm = 1.2 × 10−2 m

In case of a constructive interference, we have the relation for the distance between the two fringes as:

Where,

n = Order of fringes = 4

λ = Wavelength of light used

Hence, the wavelength of the light is 600 nm.

 

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

Speed of light, c =3 x 108 m

Frequency of incident light is given by the relation,

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 6×1014Hz. When reflected ray is normal to incident ray, the sum of the angle of incidence, ∠i and angle of reflection, ∠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:

∠i+∠r = 90°

i.e. ∠i+∠i = 90°

Hence, ∠i=902 = 45°

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

 

Let I1 and I2 be the intensity of the two light waves. Their resultant intensities can be obtained as:

Where,

= Phase difference between the two waves

For monochromatic light waves,

Phase difference =

Since path difference = λ,

Phase difference,

Given,

I’ = K

When path difference,

Phase difference,

Hence, resultant intensity,

Using equation (1), we can write:

Hence, the intensity of light at a point where the path difference is is units.

Wavelength of the light beam,

Wavelength of another light beam,

Distance of the slits from the screen = D

Distance between the two slits = d

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

(b) Let the nth bright fringe due to wavelength and (n − 1)th bright fringe due to wavelength coincide on the screen. We can equate the conditions for bright fringes as:

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

 

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

Wavelength of light used,

Angular width of the fringe in

Angular width of the fringe in water =

Refractive index of water,

Refractive index is related to angular width as:

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

 

Refractive index of glass,

Brewster angle = θ

Brewster angle is related to refractive index as:

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

 

Fresnel’s distance (ZF) is the distance for which the ray optics is a good approximation. It is given by the relation,

ZF=a2λ

Where,

Aperture width, a = 4 mm = 4 × 10-3 m

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

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