Solved Example Numerical Problems

Wave Optics - Numerical Problems Questions with Answers, Solution

Wave Nature of Light - Numerical Problems Questions with Answers, Solution

EXAMPLE Q

The wavelength of light from sodium source in vacuum is 5893Å.What are its (a) wavelength, (b) speed and (c) frequency when this light travels in water which has a refractive index of 1.33.

Solution

The refractive index of vacuum, n1 = 1

The wavelength in vacuum, λ1 = 5893 Å.

The speed in vacuum, c = 3 × 108 m s–1

The refractive index of water, n2 = 1.33

The wavelength of light in water, λ2

The speed of light in water, v2

(a) The equation relating the wavelength and refractive index is,

(b) The equation relating the speed and refractive index is,

The results show that the frequency remains same in all media.

Interference - Numerical Problems Questions with Answers, Solution

EXAMPLE Q

Two light sources with amplitudes 5 units and 3 units respectively interfere with each other. Calculate the ratio of maximum and minimum intensities.

Solution

Amplitudes, a1 = 5, a2 = 3

Resultant amplitude,

EXAMPLE Q

Two light sources of equal amplitudes interfere with each other. Calculate the ratio of maximum and minimum intensities.

Solution

Let the amplitude be a.

The intensity is, I ∝ µ 4a2 cos2(ϕ/ 2)

or I = 4I0 cos2(ϕ/2)

Resultant intensity is maximum when,

 ϕ  = 0, cos0 = 1, Imax ∝ 4a2

Resultant amplitude is minimum when,

 ϕ = π, cos(π/2) = 0, Imin = 0

Imax : Imin = 4a2 : 0

EXAMPLE Q

Two light sources have intensity of light as I0. What is the resultant intensity at a point where the two light waves have a phase difference of π/3?

Solution

Let the intensities be I0.

The resultant intensity is, I = 4I0cos2(ϕ/2)

Resultant intensity when, ϕ = π / 3, is

 I = 4I0 cos2 (π/6)

  I = 4Io (√3/2)2 = 3Io

EXAMPLE Q

The wavelength of a light is 450 nm. How much phase it will differ for a path of 3 mm?

Solution

The wavelength is, λ = 450 nm = 450×10-9m

Path difference is, δ = 3 mm = 3×10−3m

Relation between phase difference and path difference is, ϕ = 2π/λ ×δ

Substituting,

 ϕ = [ 2π / 450×10-9 ] ×3×10−3 = (π/75) ×106

 ϕ = (π/75) ×106 rad

EXAMPLE Q

In Young’s double slit experiment, the two slits are 0.15 mm apart. The light source has a wavelength of 450 nm. The screen is 2 m away from the slits.

(i) Find the distance of the second bright fringe and also third dark fringe from the central maximum.

(ii) Find the fringe width.

(iii) How will the fringe pattern change if the screen is moved away from the slits?

(iv) What will happen to the fringe width if the whole setup is immersed in water of refractive index 4/3.

Solution

d = 0.15 mm = 0.15× 10-3 m; D = 2 m;

λ = 450 nm = 450 × 10-9 m; n = 4/3

(i) Equation for nth bright fringe is,

(ii) Equation for fringe width is, 

(iii) The fringe width will increase as D is increased,

(iv) The fringe width will decrease as the setup is immersed in water of refractive index 4/3

The wavelength will decrease refractive index n times.

Hence, β ∝ λ and β’ ∝ λ’

EXAMPLE Q

Two lights of wavelengths 560 nm and 420 nm are used in Young’s double slit experiment. Find the least distance from the central fringe where the bright fringe of the two wavelengths coincides. Given D = 1 m and d = 3 mm.

Solution

λ = 560 nm = 560×10−9 m;

λ2 = 420 nm = 420×10−9 m;

D = 1 m;d = 3 mm = 3×10−3 m

For a given y, n and λ are inversely proportional.

Let nth order bright fringe of λ1 coincides with (n+1)th order bright fringe of λ2.

Thus, the 3rd bright fringe of λ1 and 4th bright fringe of λ2 coincide at the least distance y.

The least distance from the central fringe where the bright fringes of the two wavelengths coincides is, yn = n [(λD)/d]

EXAMPLE Q

Find the minimum thickness of a film of refractive index 1.25, which will strongly reflect the light of wavelength 589 nm. Also find the minimum thickness of the film to be anti-reflecting.

Solution

λ = 589 nm = 589×10−9 m

For the film to have strong reflection, the reflected waves should interfere constructively. The least optical path difference introduced by the film should be λ/2. The optical path difference between the waves reflected from the two surfaces of the film is 2µd. Thus, for strong reflection, 2µd = λ/2 [As given in equation 6.145. with n = 1]

Rewriting, d = λ/4μ

Substituting, d = [589×109] / [4×1.25] = 117.8×10−9

d = 117.8×10−9 = 117.8 nm

For the film to be anti-reflecting, the reflected rays should interfere destructively. The least optical path difference introduced by the film should be λ. The optical path difference between the waves reflected from the two surfaces of the film is 2µd. For strong reflection, 2µd = λ [As given in equation 6.146. with n = 1].

Rewriting, d = λ/2μ

Substituting, d = [589×109] / [2×1.25] = 235.6×10−9

d = 235.6×10−9 = 235.6 nm

Diffraction - Numerical Problems Questions with Answers, Solution

 EXAMPLE Q

Light of wavelength 500 nm passes through a slit of 0.2 mm wide. The diffraction pattern is formed on a screen 60 cm away. Determine the,

(i) angular spread of central maximum

(ii) the distance between the central maximum and the second minimum.

Solution

λ = 500 nm = 500×10-9 m; a = 0.2 mm = 0.2×10-3 m; D = 60 cm = 60×10-2 m

(i) Equation for diffraction minimum is, a sin θ = nλ

The central maximum is spread up to the first minimum. Hence, n = 1

(ii) To find the value of y1 for central maximum, which is spread up to first minimum with (n = 1) is, a sin θ = λ

To find the value of y2 for second minimum with (n = 2) is, a sin θ = 2λ

The distance between the central maximum and second minimum is, y2 – y1

y2 – y1 = 3 mm – 1.5 mm = 1.5 mm

Note: The above calculation shows that in the diffraction pattern caused by single slit, the width of each maximum is equal with central maximum as the double that of others. But the bright and dark fringes are not of equal width.

EXAMPLE Q

A monochromatic light of wavelength 5000 Å passes through a single slit producing diffraction pattern for the central maximum as shown in the figure. Determine the width of the slit.

Solution

λ = 5000 Å = 5000×10-10 m; sin 30º = 0.5; n = 1; a =?

Equation for diffraction minimum is, asin θ = nλ

The central maximum is spread up to the first minimum. Hence, n = 1

Rewriting, a = λ / Sinθ

Substituting, a = 5000×10−10 / 0.5

a = 1×10−6 m = 0.001×10−3m = 0.001mm

EXAMPLE Q

Calculate the distance for which ray optics is good approximation for an aperture of 5 mm and wavelength 500 nm.

Solution

a = 5 mm = 5 × 10-3 m;

λ = 500nm = 500×10−9 m; z = ?

Equation for Fresnel’s distance, z = a2/2λ

Substituting,

z = [5 × 10-3]2 / 2×500×10−9

z = 25 m