Question

In a double slit experiment, the separation between the slits is d and distance of screen from slits is D. If the wavelength of light used is `lambda` and I is the intensity of central bright fringe, then intensity at distance x from central maximum is given by ____________.

विकल्प

• `"I cos"^2 ((pi^2 "xd")/(lambda"D"))`

• `"I"^2 "sin"^2 ((pi "xd")/(2lambda"D"))`

• `"I cos"^2 ((pi "xd")/(lambda"D"))`

• `"I sin"^2 ((pi "xd")/(lambda"D"))`

Answer 1

In a double slit experiment, the separation between the slits is d and distance of screen from slits is D. If the wavelength of light used is `lambda` and I is the intensity of central bright fringe, then intensity at distance x from central maximum is given by `"I cos"^2 ((pi "xd")/(lambda"D"))`.

Explanation:

`"Applying I"_"R" = "I"_1 + "I"_2 + 2 sqrt ("I"_1 "I"_2)  "cos"  theta  "at central fringe" ("where"  theta = 0) "we get,"`

`"I"_"R" = "I"_1 + "I"_1 + 2"I"_1 = 4"I"_1`

Phase difference at a distance x when path difference becomes `"xd"/"D",` is given by

`theta' = (2pi)/lambda xx "xd"/"D"`

`therefore "I"_"R'" = "I"_1 + "I"_1 + 2"I"_1 "cos" ((2pi "xd")/(lambda"d"))`

`= "I"/4 + "I"/4 + 2"I"/4 "cos"((2pi "xd")/(lambda"d"))`

`or "I"_"R'" = "I"/2 (1 + "cos"  (2pi "cd")/(lambda "D"))`

` = "I"  "cos"^2 ((pi "xd")/(lambda "D"))`