CBSE Class 12 Physics Interference of Light Solved Examples

CBSE Class 12 Physics Interference of Light Solved Examples. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Ex.1 Consider interference between waves from two sources of Intensites I & 4I. Find intensities at points where the phase difference is π.
(A) I (B) 5I (C) 4I (D) 3I
Sol. I = R² = a₁ ² + a₂ ² + 2a₁ a₂ cos δ = I + 4I + 4I cos π I = 5I – 4I = I

Ex.2 The width of one of the two slits in a Young's double slit experiment is double of the other slit. Assuming that the amplitude of the light coming from a slit is proportional to slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.
(A) 34 : 1 (B) 9 : 1 (C) 4 : 1 (D) 16 : 1
Sol. Suppose the amplitude of the light wave coming from the narrower slit is A and that coming from the wider slit is 2A. The maximum intensity occurs at a place where constructive interference takes place. Then the resultant amplitude is the sum of the individual amplitudes. Thus, Amax = 2A + A = 3A The minimum intensity occurs at a place where destructive interference takes place. The resultant amplitude is then difference of the individual amplitudes.

Thus, A_min = 2A – A = A.

Thus, \( \frac{I_{max}}{I_{min}} = \frac{(A_{max})^2}{(A_{min})^2} = \frac{(3A)^2}{(A)^2} = 9 \).

Ex.3 The intensity of the light coming from one of the slits in a young's double slit experiment is double the intensity from the other slit. Find the ratio of the  maximum intensity to the minimum intensity in the interference fringe pattern observed.
(A) 9 : 1 (B) 34 : 1 (C) 4 : 1 (D) 16 : 1

Ex.4 The intensity ratio of two waves is 9 : 1. These waves produce the event of interference. The ratio of maximum to minimum intensity will be
(A) 1 : 9 (B) 9 : 1 (C) 1 : 4 (D) 4 : 1

Ex. 5 The equation of two light waves are

Sol. (Ex.4) \( \frac{I_1}{I_2} = \frac{9}{1} \), \( \frac{I_{max}}{I_{min}} = \left[\frac{\sqrt{I_1/I_2}+1}{\sqrt{I_1/I_2}-1}\right]^2 = \left[\frac{\sqrt{9}+1}{\sqrt{9}-1}\right]^2 = \frac{4^2}{2^2} = \frac{4}{1} \).

Sol. (Ex.5) \( a_1 = 6 \) units, \( a_2 = 8 \) units. \( \frac{I_{max}}{I_{min}} = \left[\frac{\frac{a_1}{a_2}+1}{\frac{a_1}{a_2}-1}\right]^2 = \left[\frac{\frac{6}{8}+1}{\frac{6}{8}-1}\right]^2 = \frac{49}{1} \).

Ex. 6 In a Young's slit experiment, the separation between the slits is 0.10 mm, the wavelength of light used is 600 nm and the interference pattern is observed on a screen 1.0 m away.Find the separation between the successive bright fringes.
(A) 6.6 mm (B) 6.0 mm (C) 6 m (D) 6 cm.
Sol. The separation between the sucessive bright fringes is-
β =Dλ/d
β = 6.0 mm.

Ex. 7 Two waves originating from source S1 and S2 having zero phase difference and common wavelength λ will show completely destructive interference at a point P if (S1 P – S2 P) is- [MP PMT, 87]
(A) 5λ (B) 3λ/4 (C) 2λ (D) 11λ/2

Sol. (Ex.7) For destructive interference: path difference \( S_1P - S_2P = (2n-1)\lambda/2 \). For n = 1, \( \lambda/2 \); n = 2, \( 3\lambda/2 \); n = 3, \( 5\lambda/2 \); n = 4, \( 7\lambda/2 \); n = 5, \( 9\lambda/2 \); n = 6, \( 11\lambda/2 \). So, destructive pattern is possible only for path difference = \( 11\lambda/2 \).

Ex. 8 In an interference pattern, at a point we observe the 16th order maximum for λ1=6000Aº. What order will be visible here if the source is replaced by light of wavelength λ2 = 4800 Aº.
(A) 40 (B) 20 (C) 10 (D) 80.
Sol. The distance of a bright fringe from zero order fringe is given by \( X_n = \frac{n\lambda D}{d} \). D & d are constant, so \( n_1\lambda_1 = n_2\lambda_2 \). n1 = 16, λ1 = 6000 Aº, λ2 = 4800 Aº. \( n_2 = \frac{n_1\lambda_1}{\lambda_2} = \frac{16\times 6000}{4800} = 20 \)

Ex. 9 In Young's experiment the wavelength of red light is 7.5 × 10–5 cm. and that of blue light 5.0 × 10–5 cm. The value of n for which (n + 1)th the blue bright band coincides with nth red band is-
(A) 8 (B) 4 (C) 2 (D) 1

Sol. (Ex.9) \( n_1\lambda_1 = n_2\lambda_2 \) for bright fringe: \( n(7.5\times 10^{-5}) = (n+1)(5\times 10^{-5}) \), so \( n = \frac{5.0\times 10^{-5}}{2.5\times 10^{-5}} = 2 \).

Ex. 10 In Young's double slit experiment, carried out with light of wavelength λ = 5000 Aº, the distance between the slits is 0.2 mm and the screen is at 200 cm from the slits. The central maximum is at x = 0. The third maximum will be at x equal to.
(A) 1.67 cm (B) 1.5 cm (C) 0.5 cm (D) 5.0 cm.

Sol. (Ex.10) \( X_n = \frac{n\lambda D}{d} \) or \( X_3 = \frac{3\lambda D}{d} = \frac{3\times(5000\times 10^{-8})\times 200}{0.02} = 1.5 \) cm.

Ex. 11 Two slits separated by a distance of 1mm are illieminated with red light of wavelength 6.5 × 10-7 m. The interference fringes are observed on a screen placed 1m from the slits. The distance between third dark fringe & the fifth bright fringe is equal to.
(A) .65 mm (B) 1.63 mm (C) 3.25 mm (D) 4.87 mm.
Sol. β =λD/d = 3= 6.5 ×10^-7 ×1
β = .65 × 10^–3 m = .65 mm
The distance between fifth bright fringe from third dark fringe = 2.5 β=2.5×.65 = 1.63 mm.
Ex. 12 In an experiment the two slits are 0.5 mm apart and the fringes are observed to 100 cm from the plane of the slits. The distance of the 11th bright fringe from the Ist bright fringe is 9.72 mm. Calculate the wavelength.
(A) 4.85 × 10^–5 cm (B) 4.85 × 10^–5 m (C) 4.86 × 10^–7 m (D) 4.86 × 10^–5 cm

Sol. Given d = .5 mm = 5 × 10^–2 cm, D = 100 cm. X_n = X₁₁ – X₁ = 9.72 mm. Using X_n = nλD/d, λ = X_nd/(nD) = (.972 × 5 × 10^–2)/(10 × 100) = 4.86 × 10^–5 cm.

Ex. 13 In a Young's experiment , two coherent sources are placed 0.90 mm apart and the fringes are observed one metre away. f it produces the second dark fringe at a distance of 1mm from the central fringe, the wavelength of monochromatic light used would be.
(A) 60 × 10^–4 cm (B) 10 × 10^–4 cm (C) 10 × 10^–5 cm (D) 6 × 10^–5 cm
Sol. D = 1 m, d = .90 mm = .9 × 10^–3 m. The distance of the second dark ring from centre = 10^–3 m. X_n = (2n – 1)(λ/2)(D/d). For n = 2, X_n = (3λ/2)(D/d), so λ = 2X_nd/(3D) = (2 × 10^–3 × .9 × 10^–3)/3 = 6 × 10^–7 m = 6 × 10^–5 cm.

Ex. 14 The young's double slits experiment is performed with blue and with green light of wavelength 4360 Aº and 5460 Aº respectively. It x is the distance of the 4th maxima from the central one, then
(A) x_blue = x_green (B) x_blue > x_green (C) x_blue < x_green (D)x_blue/x_green = 5460/4300
Sol. Distance of n^th maximum x = nλD/d. So x_blue/x_green = 4360/5460, hence x_blue < x_green.

Ex. 15 A beam of light consisting of two wavelength 6500 Aº & 5200 Aº is used to obtain interference fringes in a young's double slit experiment. The distance between the slits is 2.0 mm and hie distance between the plane of the slits and the screen is 120 cm. What is the least distance from the central maximum where the bright fringes due to both the wave length coincide ?
(A) .156 cm (B) .152 cm (C) .17 (D) .16 cm.
Sol. Suppose the m^th bright fringe of 6500 Aº coincides with the n^th bright fringe of 5200 Aº. Then m × 6500 = n × 5200, so m/n = 4/5. The distance y = mλ₁D/d = (4 × 6500 × 10^–10 × 1.2)/(2 × 10^–3) = .156 cm.

Ex. 16 In an experiment similar to young's experiment, interference is observed using waves associated with electrons. The electrons are being produced in an electron gun. In order to increase the fringe width.
(A) electron gun voltage be increased. (B) electron gun voltage be decreased. (C) the slit be moved away. (D) the screen be moved closer to interfering slits.
Sol. Fringe width β = λD/d. According to de Broglie, wavelength λ = h/p = h/√(2mv). As V decreases, λ increases, so β increases.

Ex. 17 Interference fringes were produced in young's double slit experiment using light of wave length 5000 Aº. When a film of material 2.5 × 10^-3 cm thick was placed over one of the slits, the fringe pettern shifted by a distance equal to 20 fringe width. The refrective inde x of the material of the film is-
(A) 1.25 (B) 1.33 (C) 1.4 (D) 1.5
Sol. Shift n = (μ – 1)tD/d. Since β = λD/d, D/d = β/λ, so n = (μ – 1)tβ/λ. With n = 20β: μ – 1 = (20 × 5000 × 10^–8)/(2.5 × 10^–3) = .4, so μ = 1.4.

Ex. 18 The path difference between two interfering waves at a point on screen is 171.5 times the wavelength. If the path difference is 0.01029 cm. Find the wavelength.
(A) 6000 × 10^–10 cm (B) 6000 Aº (C) 6000 × 10^–8 mm (D) None
Sol. Path difference = 171.5λ = (343/2)λ, an odd multiple of half wavelength, so a dark fringe is observed. So 0.01029 = (343/2)λ, giving λ = (0.01029 × 2)/343 = 6 × 10^–5 cm = 6000 Aº.

Ex. 19 Find the minimum thickness of a film which will strongly reflect the light of wavelength 598 nm. The refractive index of the material of the film is 1.25.
(A) 118 nm (B) 120 nm (C) 118 m (D) 118 mm
Sol. For strong reflection, 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, so d = λ/(4μ) = 589/(4 × 1.25) = 118 nm.

Ex.20 In young's double slit interference experiment, the distance between two sources is 0.1 / π mm.The distance of the screen from the source is 25 cm. Wavelength of light used is 5000 Aº. Then the angular position of the first dark fringe is-
(A) .10º (B) .15º (C) .30º (D) .45º
Sol. The angular position θ = β/D = λ/d (since β = λD/d). The first dark fringe will be at half the fringe width from the mid point of central maximum. Thus the angular position of first dark fringe will be α = θ/2 = (1/2)(λ/d). So α = (1/2)[(5000 × π × 10^–10)/(.1 × 10^–3)] × 180/π = .45º.

Ex. 21 In Young's double slit experiment the two slits are illuminatedby light of wavelenght 5890Aº and the distance between the fringes obtained on the screen is 0.2º. If the whole apparatus is immersed in water then the angular fringe width will be, if the refractive index of water is 4/3
(A) 0.30º (B) 0.15º (C) 15º (D) 30º
Sol. w_a = λ/d, so w_a ∝ λ. (w_a)_water/w_a = λ_water/λ = 1/μ_water. Hence (w_a)_water = (2 × 3)/4 = 0.15º.

The solved examples on interference of light are given above. CBSE Class 12 Physics Interference of Light Solved Examples