Question

A slit of width 0.59 mm is illuminated with light of wavelength 474 nm, and a screen is placed 130 cm in front of the slit. Find the widths of the first and second maxima on each side of the central maximum.

Answer 1

width of first and second minima is = 1.022 *10^{-3} m

Step-by-step explanation:

given details:

slit width =[/ex] \alpha = 0.59*10^{-9} m[/tex]

`\lambda = 474 *10^(-3) m`

distance between slit and screen =130 cm =1.30m

we know

silt distance between minima = width of the maximum

`\alpha *sin \theta = m\lambda`

distance between minima from central maxima is

`y_m = L tan\theta`

`= L sin\theta`

`= m (\lambda L)/(\alpha)`

distance between the successive minima is given as

`\Delta y =y_(m+1) - y_m`

`= (m+1) (\lambda L)/(\alpha) - m (\lambda L)/(\alpha)`

`= 1 m (\lambda L)/(\alpha)`

`= ( 474 *10^(-9)*1.30)/( 0.59*10^(-3))`

= 1.022 *10^{-3} m