Question

In a YDSE experiment two slits S₁ and S₂ have separation of d = 2 mm. The distance of the screen is D = 8/5 m. Source S starts moving from a very large distance towards S₂ perpendicular to S₁S₂ as shown in figure. The wavelength of monochromatic light is 500 nm. The number of maximas observed on the screen at point P as the source moves towards S₂ is 3995 + n. The value of n is ______.

विकल्प

• 5

• 6

• 7

• 8

Answer 1

In a YDSE experiment two slits S₁ and S₂ have separation of d = 2 mm. The distance of the screen is D = 8/5 m. Source S starts moving from a very large distance towards S₂ perpendicular to S₁S₂ as shown in figure. The wavelength of monochromatic light is 500 nm. The number of maximas observed on the screen at point P as the source moves towards S₂ is 3995 + n. The value of n is 5.

Explanation:

S₁ P – S₂ P = `"d"^2/"2D"` 

= `(2xx10^-3xx2xx0^-3)/(2xx8/5)`

= `5/2`λ

(λ = 500 nm)

As a result, because `"d"/lambda` = 4000, the first minima occurs when S is at `oo` and the last minima occurs when S is at S₂.

Therefore, there will be 4001 minima and 4000 maxima, which is equal to 3995 + 5 minima.

i.e. n = 5