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Question

In Young's Double Slit Experiment intensity at a point is (1/4) of the maximum intensity. Angular position of the point is:


A
sin1(2λd)
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B
sin1(λ2d)
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C
sin1(λ3d)
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D
sin1(λd)
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Solution

The correct option is C $$sin^{-1} (\frac{\lambda}{3d})$$
Interesting at a point $$(I)$$ where phase difference is $$\delta \ bx\ YDSE$$ in given by : 
$$I=I max \cos^{2} \delta/2$$ [$$Imax$$: maximum intensity ]
at the point where $$I=\dfrac{I max}{4}$$, we get :
$$\dfrac{I max}{4}= I max \cos^{2} \delta/2$$
$$\Rightarrow \cos^{2} \delta/2 = \dfrac{1}{4} \Rightarrow \cos \delta/2 = \pm \dfrac{\delta}{2} =\dfrac{\pi}{3}$$ or $$\dfrac{2 \pi}{3}$$
$$\Rightarrow  \delta= 2 \dfrac{\pi}{3} $$ or $$\dfrac{2 \pi}{3}$$, well take $$\dfrac{2\pi}{3}$$ only.
Thus path difference $$(\triangle x)$$ at that point in,
$$\triangle x =\dfrac{\lambda}{2 \pi} \delta = \dfrac{2 \lambda}{2 \pi}. \dfrac{\pi}{3} . \Rightarrow \triangle x =\dfrac{\lambda}{B}$$.
Also, $$\triangle x = d \sin \theta $$ ($$\theta :$$ angular position )
$$\Rightarrow \sin \theta =\dfrac{\lambda}{3d} \Rightarrow \theta = \sin^{-1} \left( \dfrac{\lambda}{3d} \right)$$

Physics

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