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)
B
sin1(λ2d)
C
sin1(λ3d)
D
sin1(λd)

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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