Monochromatic electromagnetic radiation with wavelength lambda from a distant source passes through a slit. the diffraction pattern is observed on a screen 2.50 m from the slit. a: if the width of the central maximum is 6.00 mm, what is the slit width if the wavelength is 500 nm (visible light)? b: if the width of the central maximum is 6.00 mm , what is the slit width if the wavelength is 50.0 um (infrared radiation)? c: if the width of the central maximum is 6.00 mm, what is the slit width if the wavelength is 0.500 nm (x rays)?

(a)

The formula for the diffraction from a single slit is:

where

y is distance of the nth-minimum from the centre

n is the order of the minimum

is the wavelength

D is the distance of the screen from the slit

d is the slit width

In this problem:

The width of the central maximum is 6.00 m; for n=1, y represents half the width of the central maximum, so

Solving the equation for d, we find the slit width:

(b) 0.042 m

This time, the wavelength of the radiation is

and again the value of y is

so the slit width this time is

(c)

This time, the wavelength of the x-rays is

and again the value of y is

so the slit width this time is

the results is a decrease in air temperature with an increase of altitude.

explanation: