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The intensity received at the focus (F) of the lens is I when no glass slab has been placed in front of the slit. Both the slits are of the same dimension and the plane wave front incident perpendicularly on them, has wavelength $λ .$ On placing the glass slab the intensity reduces to 3I/4 at the focus. Find out the minimum thickness of the glass slab $(i n \overset{o}{A})$ if its refractive index is 3/2. Given $λ = 6933 \dot{A}$

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answer is 2311.

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Detailed Solution

$I = 4 I_{0} c o s^{2} \frac{ϕ}{2}$

$C a s e - 1, ϕ = 0 \Rightarrow I = 4 I_{0}$

$C a s e - 2, I = \frac{3 I}{4} \Rightarrow 4 I_{0} \cos^{2} \frac{ϕ}{2} \Rightarrow \cos^{2} \frac{ϕ}{2} = \frac{3}{4} \Rightarrow \cos \frac{ϕ}{2} = \frac{\pm \sqrt{3}}{2} \Rightarrow \frac{ϕ}{2} = \frac{π}{6} \Rightarrow ϕ = \frac{π}{3}$

Now $ϕ = \frac{(μ - 1) t \times 2 π}{λ}, \frac{π}{3} = \frac{(μ - 1) t \times 2 π}{λ}, t = \frac{λ}{6 (μ - 1)} t = \frac{6933}{3} = 2311 \dot{A}$

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