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Q.

In YDSE, both slits produce equal intensities on the screen. A 100% transparent thin film is placed in front of one of the slits. Now, the intensity on the centre becomes 75% of the previous intensity. The wavelength of light is 6000 Å and refractive index of glass is 1.5. The minimum thickness of the glass slab is

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a

$0.2 μ m$

b

$0.3 μ m$

c

$0.4 μ m$

d

$0.5 μ m$

answer is A.

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Detailed Solution

$\frac{3}{4} I_{m a x} = I_{m a x} \cos^{2} \frac{ϕ}{2} \frac{ϕ}{2} = \frac{π}{6} o r ϕ = \frac{π}{3} = (\frac{2 π}{λ}) (∆ x) = (\frac{2 π}{λ}) (μ - 1) t t = \frac{λ}{6 (μ - 1)} = \frac{6000 \times 10^{- 10}}{6 (1.5 - 1)} = 0.2 \times 10^{- 6} m = 0.2 μ m$

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