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

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .