Answer to Question #92616 in Optics for ROHIT SHARMA

Question

A plane wave of wavelength 600 nm falls on a long narrow slit of width 0.6 mm.
(i) Calculate the angles of diffraction for the first two minima. (ii) How are these angles influenced if the width of slit is changed to 0.2 mm? (iii) If a convex lens of focal length 0.2 m is now placed after the slit, calculate the separation between the second minima on either side of the central maximum. (10)
b) For the arrangement described in above, repeat the calculations for the first two maxima on either side of the central maximum.

Answer 1

Diffraction minimum condition for a long narrow slit is given by formula

$$d × \sinθ = kλ (1)$$

where d is the width of the slit, θ is the angle of incidence at which the minimum intensity occurs, λ is the wavelength of the light, k=1,2,3,…

(i) In our case, λ=600 nm, d=0.6 mm, k=1

Using (1) we get

$$\sinθ=0.001$$

Answer

 θ=0.0573°

(ii)

In our case, λ=600 nm, d=0.2 mm, k=1

Using (1) we get

$$\sinθ=0.003$$

Answer

θ= 0.1719°

(iii)

Distance of nth secondary minima from central maxima is given by formula

$$y=\frac{nλf}{d} (2)$$

where f is f ocal length of converging lens

In our case, λ=600 nm, d=0.6 mm, k=2, f=0.2 m

Using (2) we get

y=0.4×10^-3 m

The separation between the second minima on either side of the central maximum is equal to 0.8×10^-3 m

Answer

0.8×10^-3 m

b)

(i) In our case, λ=600 nm, d=0.6 mm, k=2

Using (1) we get

$$\sinθ=0.002$$

Answer

θ=0.1146°

(ii)

In our case, λ=600 nm, d=0.2 mm, k=2

Using (1) we get

$$\sinθ=0.006$$

Answer

θ= 0.3438°