First slide

Young's double slit Experiment

Question

In Young's double slit experiment, the slits are of equal width and maximum intensity of light on the screen is found to be I_m. Now one of the slits is covered with a thin sheet of thickness $\frac{15 λ}{4},$ made of material of refractive index 1.2. Then intensity of light at the mid point of the source is

Moderate

A

zero
B

I_m/4
C

I_m
D

I_m/2

Solution

If I₀ be the intensity of light energy from each slit, then $I_{m} = 4 I_{0}$ .
When one of the slits is covered, path difference between rays at the mid point of the screen
$= (μ - 1) t = (1 .2 - 1) \times \frac{15 λ}{4} = \frac{3 λ}{4}$
$∴ ϕ =$ phase difference $= \frac{2 π}{λ} \times \frac{3 λ}{4} = \frac{3 π}{2}$
$∴$ At the mid point, $I = I_{m} . \cos^{2} \frac{ϕ}{2} = I_{m} \cos^{2} \frac{3 π}{4} = I_{m} {(\frac{I}{V_{2}})}^{2} = \frac{I_{m}}{2}$

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