Question

In Young’s double slit experiment, d = 1mm, $\lambda = 6000\mathop {\text{A}}\limits^0 $ & D = 1 m. The slits produce the same intensity on the screen. The minimum distance between two points on the screen having 75% intensity of the maximum intensity is:
(A). 0.45 mm
(B). 0.40 mm
(C). 0.30 mm
(D). 0.20 mm

Answer 1

Hint: The central maximum has maximum intensity and this intensity decreases as we go away from the centre. The decrease in intensity depends on the cosine square of half the phase difference between the incident waves of light.

Complete step by step solution:
Interference of light: When two light waves from two coherent sources overlap with each other, then they superimpose to form regions of light and darkness. The points where constructive interference takes place, we get maximum light whereas points of destructive interference are dark.
The coherent sources of light are those which have the same frequency and constant phase difference. Such sources can be obtained from a single source by passing the light from that source through two slits. Young’s double slit experiment uses this type of arrangement. The pattern of maximum light and minimum light is called fringe pattern.
We are given that:
Slit width: d = 1mm = ${10^{ - 3}}m$
Wavelength of light: $\lambda = 6000\mathop {\text{A}}\limits^0 = 6 \times {10^{ - 7}}m$
Distance between slits and screen: D = 1m
$\dfrac{I}{{{I_0}}} = \dfrac{{75}}{{100}} = \dfrac{3}{4}$
Intensity of light in the fringe pattern is given as $I = {I_0}{\cos ^2}\left( {\dfrac{\varphi }{2}} \right)$
 where ${I_0}$ is the maximum intensity and $\varphi $ is phase difference; $\varphi = \dfrac{{2\pi }}{\lambda }\Delta x$ where $\Delta x$ is the path difference; $\Delta x = \dfrac{{yd}}{D}$ where y is the distance between the two maxima.
Substituting various values, we get
$ I = {I_0}{\cos ^2}\left( {\dfrac{{\pi yd}}{{\lambda D}}} \right) \\ $
$ \Rightarrow \dfrac{3}{4} = {\cos ^2}\left( {\dfrac{{\pi yd}}{{\lambda D}}} \right) \\ $
$ \Rightarrow {\cos ^2}\left( {\dfrac{\pi }{6}} \right) = {\cos ^2}\left( {\dfrac{{\pi yd}}{{\lambda D}}} \right) \\ $
  $ \Rightarrow \dfrac{{\pi yd}}{{\lambda D}} = \dfrac{\pi }{6} \\ $
  $ \Rightarrow y = \dfrac{{\lambda D}}{{6d}} \\ $
  $ \Rightarrow y = \dfrac{{6 \times {{10}^{ - 7}} \times 1}}{{6 \times {{10}^{ - 3}}}} \\ $
  $ \Rightarrow y = {10^{ - 4}}m = 0.1mm \\ $
Shortest distance between two maxima having 75% intensity = 2y = 0.2mm

Note: y is the distance between central maxima and the maxima having 75% intensity. The pattern is symmetric about the central maxima so the distance between the two maxima with 75% intensity on either side of the central maxima is 2y.