Question

Light of wavelength 588.0 nm illuminates a slit of width 0.68 mm. (a) At what distance from the slit should a screen be placed if the first minimum in the diffraction pattern is to be 0.89 mm from the central maximum? m (b) Calculate the width of the central maximum. mm

Answer 1

(a) 1.029 m

(b) 1.78 mm

Solution:

As per the question:

Wavelength of light,
`\lambda = 588\ nm`

Slit width, w = 0.68 mm

Now,

The distance first minimum in the diffraction pattern,
`y_(1) = 0.89\ mm`

Now,

(a) For first minima:

`wsin\theta = \lambda`

`w((y)/(D)) = \lambda`

where

D = Distance from the screen

`D = ((0.68* 10^(- 3)* 0.89* 10^(- 3))/(588* 10^(- 9)))`

D = 1.029 m

(b) The width of the central maxima is given by:

2y =
`2* 0.89 = 1.78\ mm`