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Q.

In YDSE , $D$ = 1.2 m and $d$ = 0.25 cm, the slits are illuminated with coherent 600 nm light. Calculate the distance $y$ above the central maximum for which the average intensity on the screen is 75% of the maximum.

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a

78 $μ m$

b

48 $μ m$

c

39 $μ m$

d

56 $μ m$

answer is D.

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Detailed Solution

$I = I_{m a x} \cos^{2} \frac{ϕ}{2} \frac{3}{4} I_{m a x} = I_{m a x} \cos^{2} \frac{ϕ}{2} o r \cos \frac{ϕ}{2} = \frac{\sqrt{3}}{2} t h e r e f o r e, \frac{ϕ}{2} = \frac{π}{6} ϕ = \frac{π}{3} \Rightarrow (\frac{2 π}{λ}) (∆ x) \Rightarrow (\frac{2 π}{λ}) (\frac{y d}{D}) y = \frac{λ D}{6 d} \Rightarrow \frac{(600 \times 10^{- 19}) (1.2)}{(6) (0.25 \times 10^{- 2})} \Rightarrow 48 \times 10^{- 6} m \Rightarrow 48 μ m$

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