Question

In an interference experiment, the intensity at a point is `(1/4)^"th"` of the maximum intensity. The angular position of this point is at ____________. 
(cos 60° = 0.5, `lambda` = wavelength of light, d = slit width)

विकल्प

• sin^-1 (λ/4d)

• sin^-1 (λ/2d)

• sin^-1 (λ/d)

• sin^-1 (λ/3d)

Answer 1

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

Explanation:

Maximum intensity I_max = 4I₀

where I₀ is the intensity of each wave.

`I_"max"/4 = I_0`

I = `4I_0 cos^2  phi/2`

where I = I₀, we have `I_0 = 4I_0 cos^2  phi/2`

`cos^2  phi/2 = 1/4` or `cos  phi/2 = 1/2`

∴ `phi/2 = pi/3` or `phi = (2pi)/3`

For one fringe width X = `(lambdaD)/d`

Angular separation = `x/D = lambda/d`

For phase difference of 2π, the angular separation is given by

sinθ ≈ θ = `lambda/d`

For phase difference of `(2pi)/3`, the angular separation is given by `sintheta = lambda/(3d)` or `theta = sin^-1(lambda/(3d))`