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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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प्रश्न

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.

योग

उत्तर

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 × 10-3 m

To find: Distance between slits (d) 

Formula: W = `(λ  "D")/"d"`

Calculation:

From formula,

`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 xx 10^-4`m

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

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