Question
In Young's double slit experiment, a fine beam of coherent monochromatic light of wavelength λ = 600 nm is incident on identical slits S1 and S2 at separation d. The intensity at the central maximum formed at O is Imax and the angular fringe width is β = 0.1°. When a thin transparent film is placed in front of the slit S2, the intensity at O changes. It is found that the smallest thickness of the film for which the intensity at O becomes half the maximum intensity (i.e. Imax/2) is 250 nm. Neglecting the absorption of light by the film, the zero order fringe earlier at O now forms at O′ where OO′ = 0.5 mm. Choose the correct options:-
Options
Solution
[Key Concept: A film of thickness t adds optical path (μ − 1)t and shifts the whole pattern towards the covered slit.]
$$\lambda = 600n\text{ }mI_{\text{max~}} = 4I_{0}\ \beta = \frac{\lambda}{d} = {0.1}^{\circ}$$
$$ { \begin{matrix} & I = 4I_{0}\cos^{2}\frac{\theta}{2} \\ & 2I_{0} = 4I_{0}\cos^{2}\frac{\theta}{2} \\ & cos\frac{\theta}{2} = \pm \frac{1}{\sqrt{2}} \\ & \ \therefore\theta = \frac{\pi}{2} = \frac{2\pi}{\lambda}\Delta x \\ & \Delta x = \frac{\lambda}{4} \\ & \ (\mu - 1)t = \frac{\lambda}{4} \\ & \mu - 1 = \frac{1}{4} \times \frac{600 \times 10^{- 9}}{250 \times 10^{- 9}} = 0.6 \\ & \ \therefore\mu = 1.6\ \text{~Option (a)~} \\ & OO^{'} = dsin\theta \approx \text{~dtan~}\theta = \frac{\lambda}{4} \\ & \text{ }d = \frac{0.5 \times 10^{- 3}}{D} = \frac{\lambda}{4} \\ & \omega = \frac{\lambda D}{\text{ }d} = \text{~fringe width~} = 4 \times 0.5 \times 10^{- 3} = 2\text{ }mm \\ & \omega = 2\text{ }mm\ \text{~Option~}(b) \\ & \omega = \beta D \\ & D = \frac{\omega}{\beta} = \frac{2 \times 10^{- 3}}{0.1 \times \pi} \times 180 \approx 1.15\text{ }m\ \text{~Option (d)~} \end{matrix}}$$.
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