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A Plane Wavefront of light of wavelength 5500 A.U. is incident on two slits in a screen perpendicular to the direction of light rays. If the total separation - Physics

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Question

A Plane Wavefront of light of wavelength 5500 A.U. is incident on two slits in a screen perpendicular to the direction of light rays. If the total separation of 10 bright fringes on a screen 2 m away is 2 cm. Find the distance between the slits.

Numerical
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Solution

Given: λ = 5500 A.U. = 5500 × 10−10 m

D = 2 m 

Distance between 10 fringes = 2 cm = 0.02 m

Fringe width W = `0.02/10` = 0.002 m = 2 × 103 m

To find: Distance between slits (d) = ?

∴ Fringe width (W) = `(λ D)/d`

`2 xx 10^-3 = (5500 xx 10^-10 xx 2)/d`

∴ d = `(5.5 xx 10^-7 xx 2)/(2 xx 10^-3)`

= 5.5 × 10–4 m

The distance between two slits is 5.5 × 10-4 m.

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Chapter 7: Wave Optics - Short Answer I

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