Question

In a YDSE a parallel beam of light of wavelength $6000 \overset{°}{A}$ is incident on slits at angle of incidence $30 °$ . $A$ and $B$ are two thin transparent films each of refractive index $1.5$ . Thickness of $A$ is $20.4 μm$ . Light coming through $A$ and $B$ have intensities $I$ and $4 I$ respectively on the screen. Intensity at point $O$ which is symmetric relative to the slits is $3 I$ . The central maxima is above $O$ .
(i) What is the maximum thickness of $B$ to do so. Assuming thickness of $B$ to be that found in part (i) answer the following parts.

(ii) Find fringe width, maximum intensity and minimum intensity on screen.

(iii) Distance of the nearest minima from $O$ .

(iv) Intensity at $5 cm$ on either side of $O$ .

Accepted Answer

Given: The wavelength $λ = 6000 \overset{°}{A}$ , angle of incident $A$ , refractive index of plates $μ = 1.5$ , thickness of $A$ is $t_{A} = 20.4 μ m$ and separation between the slit and screen is $D = 1 m$ .

(i) Resultant intensity of the interference pattern is given as,

$I_{res} = I_{1} + I_{2} + 2 \sqrt{I_{1}} \sqrt{I_{2}} \cos ϕ$ ; where $ϕ$ is phase difference.

$3 I = 4 I + I + 2 \sqrt{4 I} \sqrt{I} \cos ϕ$

$\cos ϕ = \frac{- 1}{2} \Rightarrow ϕ = \frac{2 π}{3}$

This means path difference between the waves is,

$∆ x = \frac{ϕ}{(\frac{2 π}{λ})} = \frac{(\frac{2 π}{3})}{(\frac{2 π}{λ})} = \frac{λ}{3} = 2 \times 10^{- 7} m$ .

Hence, the sum of the initial path difference between the waves must be equal to the path difference between the waves due to glass slabs, i.e.,

$d \sin θ + (μ - 1) t_{A} - (μ - 1) t_{B} = ∆ x$

$10^{- 4} \times \sin 30 ° + (1.5 - 1) 20.4 \times 10^{- 6} - (1.5 - 1) t_{B} = 2 \times 10^{- 7}$

$500 \times 10^{- 7} + 102 \times 10^{- 7} - (0.5) t_{B} = 2 \times 10^{- 7}$

$t_{B} = \frac{(500 + 100) \times 10^{- 7}}{0.5} = 1200 \times 10^{- 7} = 120 μm$

(ii). Fringe width of interference pattern is given as,

$β = \frac{λ D}{d} = \frac{6000 \times 10^{- 10} \times 1}{10^{- 4}} = 6 mm$ .

(iii) The Nearest minima will be obtained at phase difference of $π$ by moving upwards from $O$ .

Extra change in phase required for nearest minima is $π - \frac{2 π}{3} = \frac{π}{3}$ .

Extra path difference due to given phase is $∆ x = \frac{λ}{2 π} \times \frac{π}{3} = \frac{λ}{6}$ .

For minima, $∆ x = \frac{λ}{6} = \frac{2 n + 1}{2} λ \Rightarrow \frac{1}{6} = \frac{2 n + 1}{2}$ .

Distance from $O$ is $y = \frac{2 n + 1}{2} β = \frac{6 mm}{6} = 1 mm$ .

(iv) Calculating for upwards side,

Number of fringes formed in $5 cm$ is $\frac{50}{β} = \frac{50}{6} = 8 + \frac{1}{3}$ .

Hence, the net variation in intensity will be due to ${(\frac{1}{3})}^{rd}$ portion of fringe width only. Path difference between the waves at $\frac{β}{3}$ is $\frac{λ}{3}$ .

Phase difference at that point is $ϕ = \frac{2 π}{λ} \times \frac{λ}{3} = \frac{2 π}{3}$ .

There was an initial phase difference of $\frac{2 π}{3}$ between the waves at $O$ . Hence, total phase difference is $0$ .

Hence, intensity is given as $I_{res} = 4 I + I + 2 \sqrt{4 I} \sqrt{I} \cos (0) = 9 I$

Similarly, if we downwards, phase difference due to position on screen will be same, i.e., $\frac{2 π}{3}$ . But this time the phases will not cancel each other rather gets added up. Net phase difference is $\frac{4 π}{3}$ .

Hence, intensity is given as $I_{res} = 4 I + I + 2 \sqrt{4 I} \sqrt{I} \cos (\frac{4 π}{3}) = 3 I$ .

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