Question

A single slit diffraction pasttern has the central maximum width of 1 cm. Given the wavelength of 500 nm and slit width of 0.01 mm, what is the distance from the slit to the observation screen? 1 m

0.1 m
71 cm
0.71 m
none of the above?

Answer 1

distance between slit and screen, d = 0.1 m

Given:

slit width, x =
`0.01 mm = 0.01* 10^(-3) m`

width of central maximum, t = 1 cm

wavelength,
`\lambda = 500nm = 500* 10^(-9) m`

Step-by-step explanation:

The first minima will be at the position t:

t =
`(1)/(2) = (1)/(2) = 0.5 cm`

Now, using the formula:

xsin
`\theta = n\lambda`

where

n = 1

xsin
`\theta = \lambda`

sin
`\theta = (\lambda)/(x)`

sin
`\theta = (500* 10^(-9))/( 0.01* 10^(-3)) = 0.05`

Now, calculation of distance between slit and screen, d:

`d = (t)/(sin\theta)`

`d = (0.5* 10^(-2))/(sin0.05) = 0.1 m`