JEE Main 2014 · 12 Apr · Shift 1 · Physics Q24

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PhysicsSingle correct+4 / -1Question 24 of 90

Interference pattern is observed at 'P' due to superimposition of two rays coming out from a source 'S' as shown in the figure. The value of ' $l$ ' for which maxima is obtained at 'P' is :

(R is perfect reflecting surface) :

S and P are a distance 2l apart on a horizontal line; one ray goes directly from S to P, the other goes down to the mirror R below, meeting it at 30° midway between S and P, and reflects up to P.

l=\dfrac{2n\lambda}{\sqrt3-1}

l=\dfrac{(2n-1)\lambda}{2(\sqrt3-1)}

l=\dfrac{(2n-1)\lambda\sqrt3}{4(2-\sqrt3)}

l=\dfrac{(2n-1)\lambda}{\sqrt3-1}

Answer and worked solution

Official answer: C

By symmetry the reflected ray meets R midway, at 30° to the surface. Each leg is $\dfrac l{\cos30^\circ}=\dfrac{2l}{\sqrt3}$ , so the path difference is
$\frac{4l}{\sqrt3}-2l=\frac{2l(2-\sqrt3)}{\sqrt3}.$
Reflection at R adds $\frac\lambda2$ , so maxima need this to be $(2n-1)\frac\lambda2$ :
$l=\frac{(2n-1)\lambda\sqrt3}{4(2-\sqrt3)}.$

Topics:Interference and Young's double slit Reflection and plane mirrors

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JEE Main 2014 · 12 Apr · Shift 1, Physics question 24