In a standard YDSE setup, the fringe width on the screen is 1.5mm. When a thin glass film is pasted in front of the upper slit, the fringe pattern shifts up. But it is seen that a point P above central maxima where intensity was one fourth the intensity at central maxima, intensity remains the same. There were no maxima between central maxima and point P before film was introduced. What can be the thickness of the film? Take $μ = 15, λ = 450 n m$ :-

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$5 \times 10^{- 7} m$

$1.4 \times 10^{- 6} m$

$9 \times 10^{- 7} m$

$3 \times 10^{- 7} m$

answer is A, C.

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Detailed Solution

$I_{0} = 4 I_{0} \cos^{2} \frac{π y d}{λ D} \cos \frac{π y d}{λ D} = \pm \frac{1}{2} \frac{π y d}{λ D} = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3} ........... y = \frac{β}{3}, \frac{2 β}{3}, \frac{4 β}{3} ......... Δ y = \frac{β}{3} o r β o r \frac{2 β}{3} o r \frac{4 β}{3}$