Question

In a Young's double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is $\dfrac{1}{8}$ th of wavelength. What is the ratio of the intensity at this point to that at the center of a bright fringe is close to?
(A) 0.94
(B) 0.74
(C) 0.85
(D) 0.80

Answer 1

Hint:In order to solve this problem,we are going to apply the concept of intensity distribution in a Young's double slit experiment,concept of path difference and phase difference.If the intensity of the central bright fringe is ${I_o}$, the intensity at a point is given by the formula, $I = {I_o}{\cos ^2}\left( {\dfrac{{\Delta \phi }}{2}} \right)$, where $\Delta \phi $ is the phase difference between the two waves.

Complete step by step answer:
The path difference between the two waves is given to be $\dfrac{1}{8}$th of the wavelength of the interfering waves. It can be written as,
$\Delta x = \dfrac{\lambda }{8}$, where $\lambda $ is the wavelength of the interfering waves.
To find the intensity at the given point with respect to the intensity of the central bright fringe, we need to find the phase difference. It can be found out by using the relation between path difference and phase difference, which is,
$\Delta \phi = \dfrac{{2\pi \Delta x}}{\lambda }$ …equation (1)
We now substitute the value of path difference in terms of wavelength in equation (1). On doing that, we obtain,
$\Delta \phi = \dfrac{{2\pi }}{\lambda } \times \dfrac{\lambda }{8} = \dfrac{\pi }{4}$
Since we have obtained the value of the phase difference, we can use it to find out the intensity at the given point in terms of the intensity of the central bright fringe with the help of the following equation,
$I = {I_o}{\cos ^2}\left( {\dfrac{{\Delta \phi }}{2}} \right)$ …equation (2)
Substituting the value of phase difference in equation (2), we obtain,
$I = {I_o}{\cos ^2}\left( {\dfrac{\pi }{8}} \right)\\
\Rightarrow \dfrac{I}{{{I_o}}} = {\cos ^2}\left( {\dfrac{\pi }{8}} \right) \\
\therefore \dfrac{I}{{{I_o}}}= 0.85$

Hence, the ratio of intensity is 0.85.Therefore, the correct answer is option C.

Note: In the given question, the intensity of the point on the screen does not depend on the wavelength of the light. This is because the path difference was written in terms of the wavelength. If the path difference would have been mentioned as a numerical value, we would have required the value of the wavelength of the light.