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

Answer:

(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:

$wsin\theta = \lambda$

$w(\frac{y}{D}) = \lambda$

where

D = Distance from the screen

$D = (\frac{0.68\times 10^{- 3}\times 0.89\times 10^{- 3}}{588\times 10^{- 9}})$

D = 1.029 m

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

2y = $2\times 0.89 = 1.78\ mm$