Sol. When a monochromatic light of wavelength λ is used to obtain interference pattern in Young’s double slit experiment, fringe width is given by ω = where D is distance of screen from slits and d is distance between the slits. Hence, fringe width for light of wavelength λ 1 , ω 1 = ∴ ω 1 = 80 µm

and fringe width for light of wavelength, λ 2 , ω 2 = = 112 µm Since, the incident light beam has both the wavelengths λ 1 and λ 2 , therefore, interference patterns are formed on the screen for both the wavelengths. A black line is formed at the position where dark fringes are formed for both of the wavelengths.

Let first black line be formed at distance y from central bright fringe. Let at this position there be mth dark fringe of wavelength λ 1 and nth dark fringe of wavelength λ 2 .

∴ Distance of first black line, from central bright line,

y = ω 1 = ω 2 … (1)

or = … (2)

For first black line, y should be minimum possible which corresponds to least possible integer values of m and n.

Hence, = or m = 4, n = 3

∴ Position of first black line

y = ω 1 = 280 µm Ans. (i)

Since, interference pattern is always symmetric about central bright fringe, therefore, there are two first black lines one is at height y from central bright fringe and the other at a depth y from it.

Hence, distance between two consecutive black lines = 2y = 560 µm Ans. (ii)