Here is a question on Young’s double slit which appeared in Kerala Medical Entrance 2007 question paper:
In a two slit experiment, with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5×10–2 m towards the slits, the change in fringe width is 10–3 m. Then the wave length of light used is (given that distance between the slits is 0.03 mm)
(a) 4000 Ǻ (b) 4500 Ǻ (c) 5000 Ǻ (d) 5500 Ǻ (e) 6000 Ǻ
It is enough to write two equations for the fringe width and solve for the wave length λ. If β is the initial fringe width, we have
β = λD/d where ‘D’ is the initial distance of the screen from the double slit and ‘d’ is the distance between the slits.
When D is decreased, β is decreased. Therefore, when the screen is moved by 5×10–2 m towards the slits, we have
β –10–3 = λ(D – 5×10–2 )/d
Subtracting this equation from the first one, we obtain
10–3 = λ×(5×10–2)/d
Since d = 0.03 mm = 0.03×10–6 m,
λ = (0.03×10–6)/(5×10–2) m = 6000×10–10 m = 6000 Ǻ. [Option (e)].
