Question

A monochromatic light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen. Estimate the number of fringes obtained in young's double slit experiment with fringe width 0.5 mm, which can be accommodated within the region of total angular spread of the central maximum due to single slit.

Answer 1

Hey dear,

● Answer -
Width of central maximum = 5×10^-3
No of fringes obtained = 5 fringes

● Explanation -
# Given -
a = 0.2 mm = 2×10^-4 m
λ = 500 nm = 5×10^-7 m

# Solution -
Angular width of central maximum -
θ = 2λ / a
θ = 2×5×10^-7 / 2×10^-4
θ = 5×10^-3 rad

No of fringes that can be accommodated within spread of central maximum -
n = 2d / a
n = 2×5×10^-4 / 2×10^-4
n = 5 fringes

Therefore, angular width of central maximum is 5×10^-3 rad and 5 fringes will be accommodated within that.

Hope this helps you...