Question

Monochromatic light of wavelength 630 nm is incident on a slit of width 16 µm. A diffraction pattern is produced on a screen 2.0 m away from the slit.
What is the width of the central peak in the diffraction pattern produced on the screen? (to 2 s.f and in cm)

Answer 1

The width of the central peak b is given by formula

$$b=\frac { 2Lkλ}{a} (1)$$

where k=1, L=2.0 m, λ=630 nm, a=16 µm

Using (1) we get:

b=157.5×10^-3 m

Answer:

157.5×10^-3 m