Question

The maximum intensity of fringes in Young's experiment is I. If one of its slits is closed, the intensity at that place becomes Io. Then the relationship between I and Io is
A) $ I = {I_o} $ $ $
B) $ I = 2{I_o} $
C) $ I = 4{I_o} $
D) There is no relationship between $ I $ and $ {I_o} $

Answer 1

Hint: According to the question, we need to find the relationship between I (maximum intensity of fringes in young's experiment) and I (intensity when one slit is closed). So, here we will solve this question using the results from the young’s experiment, then we will put the appropriate values and this will give us the result.
 $ {I_R} = 4{I'}{\cos ^2}\phi $ , where $ \phi $ is the phase difference between the waves at the point of observation.

Complete step by step answer:
Young’s double-slit experiment uses two coherent sources of light placed at a small distance apart, usually, only a few orders of magnitude greater than the wavelength of light is used. Young’s double-slit experiment helped in understanding the wave theory of light which is explained with the help of a diagram. A screen or photo detector is placed at a large distance ’D’ away from the slits. Thomas Young's experiment with light was part of classical physics long before the development of quantum mechanics and the concept of wave-particle duality. He believed it demonstrated that the wave theory of light was correct, and his experiment is sometimes referred to as Young’s experiment or Young's slits.
Let us suppose slit widths are equal so they produces waves of equal intensity say I′, Resultant intensity at any point,
 $ {I_R} = 4{I'}{\cos^2}\phi $ , where $ \phi $ is the phase difference between the waves at the point of observation. We know that for maximum intensity, $ \phi = {0^ \circ } $
Putting the value of $ \phi $ in $ {I_R} = 4{I'}{\cos ^2}\phi $ , we get,
 $ {I_{\max }} = 4I' = I......(1) $
Now if one of the slits is closed. Resultant intensity at the same point will be $ I' $ only, that is $ I' = {I_o}......(2) $
Now by comparing equation (1) and equation (2), we get, $ I = 4{I_o} $
So the final answer is option (C) $ I = 4{I_o} $ .

Note:
Students should keep in mind the units. Calculations should be done in standard units, otherwise it will give the wrong answer. Formulas for maximum intensity of the fringes in Young's experiment must be remembered in order to solve the problem and get to the final answer. Further, it is advisable for the students to remember the values of trigonometric functions.