In Young's double slit experiment intensity at a point is of the maximum intensity. Angular position of this point is [2005]

Question

In Young's double slit experiment intensity at a point is $(\frac{1}{4})$ of the maximum intensity. Angular position of this point is [2005]

Smart Achievers · Accepted Answer

(3)

$I = I_{\max} \cos^{2}$

$= I_{\max} \cos^{2} (\frac{π d \sin θ}{λ})$

$4 = \frac{2 π}{λ} d \sin θ$

$\Rightarrow \frac{I_{\max}}{4} = I_{\max} \cos^{2} (\frac{π d \sin θ}{λ})$

or, $\cos (\frac{π d \sin θ}{λ}) = \frac{1}{2}$ $\Rightarrow \frac{π d \sin θ}{λ} = \frac{π}{3}$

$∴$ $θ = \sin^{- 1} (\frac{λ}{3 d})$

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