Light of $\lambda$ = 5000 $\mathring{A}$ falls normally on a single slit of width 0.3 mm. Calculate the total angular as well as linear width of the central maximum obtained on the screen placed at 1 m from the slits.

Subject: Applied Physics 2

Topic: Interference And Diffraction

Difficulty: Medium

λ = 5000 x 10$^{-8}$ cm, a= 3 x10$^{-2}$ cm,first minimum so n=1

Distance of screen from the slit 100 cm = f = 100 cm

a sinθ = nλ

Total angular width of the central maximum is twice the angle made by the first minimum with the normal to the slit

$ sin \ θ = \frac{1 \times 5000 \times 10^{-8}}{3 \times 10^{-2}} = 1.6 \times 10^{-3} $

θ = 9.2

Total angular width is 2θ = 2x 9.2 =18.4

The linear width of the central maximum on the screen is 2x where $ x= \frac{fλ}{a} = \frac{100 \times 5000 \times 10^{-8}}{3 \times 10^{-2}}$

Therefore x= 0.16

So linear width is 2 x 0.16 = 0.33 cm