Question

In a Young’s double slit experiment the intensity at a point where the path difference is $\dfrac{\lambda }{6}$ ( $\lambda$ being the wavelength of the light used) is $I$. If ${I_0}$ denotes the maximum intensity, $\dfrac{I}{{{I_0}}}$ is equal to
(A) $\dfrac{{\sqrt 3 }}{2}$
(B) $\dfrac{1}{{\sqrt 2 }}$
(C) $\dfrac{3}{4}$
(D) $\dfrac{1}{2}$

Answer 1

Hint
To solve this question, we need to obtain the value of the phase difference from the given value of the path difference. Then, using the relation of the intensity with the phase difference, we will get the required ratio.
Formula Used: The formulae which are used in solving this question are given by
$\Rightarrow \varphi = \dfrac{{2\pi }}{\lambda }\Delta x$, here $\varphi$ is the phase difference corresponding to the path difference of $\Delta x$ and $\lambda$ is the wavelength of the light.
$\Rightarrow I = {I_0}{\cos ^2}\left( {\dfrac{\varphi }{2}} \right)$, here $I$ is the intensity of light at a point, where the phase difference is $\varphi$, and ${I_0}$ is the maximum intensity for the case of Young’s double slit experiment.

Complete step by step answer
Let the amplitude be $A$
We know that for a path difference of $\lambda$ the phase difference is $2\pi$
So for a path difference of $\Delta x$, phase difference $\varphi$ is given by
$\Rightarrow \varphi = \dfrac{{2\pi }}{\lambda }\Delta x$ …...(1)
According to the question the path difference is $\dfrac{\lambda }{6}$
So we have
$\Rightarrow \Delta x = \dfrac{\lambda }{6}$
Putting in (1) we get the corresponding phase difference as
$\Rightarrow \varphi = \dfrac{{2\pi }}{\lambda }\left( {\dfrac{\lambda }{6}} \right)$
On simplifying we get
$\Rightarrow \varphi = \dfrac{\pi }{3}$ ……..(2)
Now, the intensity is related to the phase difference by the formula,
$\Rightarrow I = {I_0}{\cos ^2}\left( {\dfrac{\varphi }{2}} \right)$
From (2) on substituting we get
$\Rightarrow I = {I_0}{\cos ^2}\left( {\dfrac{\pi }{6}} \right)$
$\Rightarrow I = \dfrac{3}{4}{I_0}$
Dividing by ${I_0}$ on both the sides, we get
$\Rightarrow \dfrac{I}{{{I_0}}} = \dfrac{3}{4}$
Hence, the correct answer is option (C).

Note
We should not get confused by the formula of the intensity which is given in the form of the intensity produced by the individual sources at each of the two slits. That formula is given by the relation
$\Rightarrow I = 4{I_0}{\cos ^2}\left( {\dfrac{\varphi }{2}} \right)$
We must note that the maximum intensity given in this question is not the intensity produced by the individual sources. It is the maximum value of the resultant intensity, which is produced at the screen. The $4{I_0}$ term in the above equation is taken to be equal to the maximum value of the resultant intensity. So we must be very much careful regarding this fact.