A0.50 mm-wide slit is illuminated by light of wavelength 500 nm.what is the width of the central maximum on a screen 2.0m behind the slit?

0.004 m

Explanation:

For light passing through a single slit, the position of the nth-minimum in the diffraction pattern is given by

where

is the wavelength

D is the distance of the screen from the slit

d is the width of the slit

Therefore, the width of the central maximum is equal to twice the value of y for n=1 (first minimum):

where we have

is the wavelength

D = 2.0 m is the distance of the screen

is the width of the slit

Substituting, we find

Refractive index(n)= speed of light in vacuum/ speed of light in medium.
Since refractive index is 1.11, and usually we know that the speed of light is 300,000km/s, simply divide 300,000km/s over 1.11. That is, 270,270.2703 when simply rounding, its 270,270 km/s.