Question

In interference experiment, intensity at a point is `(1/4)^"th"` of the maximum intensity. The angular position of this point is at (sin30° = cos60° = 0.5, `lambda` = wavelength of light, d = slit width) ____________.

Options

• sin^-1 (`lambda`/3d)

• sin^-1 (`lambda`/2d)

• sin^-1 (`lambda`/4d)

• sin^-1 (`lambda`/d)

Answer 1

In interference experiment, intensity at a point is `(1/4)^"th"` of the maximum intensity. The angular position of this point is at sin^-1 (`lambda`/3d).

Explanation:

For any point in interference pattern,

`"I" = "I"_"max" "cos"^2  phi/2`

`therefore "I"_"max"/4 = "I"_"max" "cos"^2  phi/2`

`therefore "cos"^2  phi/2 = 1/4`

`therefore "cos"^2  phi/2 = 1/2`

`therefore phi/2 = 60° = pi/3`

`therefore phi = (2pi)/3`

We know that,

`phi = ((2pi)/lambda) Delta"x,"  "where"  Delta"x"` is path difference

and `Delta "x" = "d sin" theta`

`therefore (2pi)/3 = (2pi)/lambda ("d sin" theta)`

`therefore lambda/(3"d") = "sin" theta`

`therefore theta = "sin"^-1 (lambda/(3"d"))`