First slide

Young's double slit Experiment

Question

In young's double slit experiment, intensity of central maximum is $I_{m}$ . If wavelength of light used is $6000 \overset{o}{A}$ , find the path difference between two waves emerging from the slits at a point on the screen where the intensity of light is $I_{m} / 4$ .

Easy

A

1000 $\overset{o}{A}$
B

4000 $\overset{o}{A}$
C

2000 $\overset{o}{A}$
D

3000 $\overset{o}{A}$

Solution

$I = I_{m} \cos^{2} \frac{ϕ}{2} \Rightarrow \frac{I_{m}}{4} = I_{m} \cos^{2} \frac{ϕ}{2}$
$\Rightarrow \cos \frac{ϕ}{2} = \frac{1}{2} \Rightarrow \frac{ϕ}{2} = \frac{π}{3} \Rightarrow ϕ = \frac{2 π}{3}$
Now $\frac{2 π}{λ} \times p a t h d i f f e r e n c e = ϕ \Rightarrow \frac{2 π}{6000} \times p a t h d i f f e r e n c e = \frac{2 π}{3} \Rightarrow p a t h d i f f e r e n c e = 2000 \overset{o}{A}$

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