Question

Light of wavelength 600 nm is incident on a pair of slits 2,000 nm apart. Find the angular deviation of (a) the first minimum, (b) the first, second, and third maxima above the central one. (c) What is the longest wavelength for which there are four maxima above the central one? (d) The same light is incident on a diffraction grating with adjacent slits 2,000 nm apart. Compare and contrast the resulting interference pattern with that of the two-slit system.

Answer 1

(a)

d sin = (m+0.5 ) For minima

d = slit width = 2000 X 10^(-9) Mrs   = 600 X 10^(-9) mrs

For 1st minimum m= 0

2000 X 10^(-9) Mrs X sin = 600 X 10^(-9) mrs/2

Sin = (3/10) (1/2) = 3/20

= 8.63 deg

(b) dsin = m

first max 2000 X 10^(-9) X sin = (1) 600 X 10^(-9)

Sin1 = (3/10)

1 = 17.5 deg

2000 X 10^(-9) X sin 2 = (2) X 600 X 10^(-9)

2 = 36.9 deg

2000 X 10^(-9) X sin 3 = (3) X 600 X 10^(-9)

3 = 64.1 deg

(c) to find longest wavelength = 90

2000 X 10^(-9) x Sin 90 = 5 X

= 400 nm

(d) If grating with d = 2000 nm is used the position of maxima and minima does not change but the diffraction pattern is sharper.