Question

If a Young’s double slit experiment with light of wavelength $\lambda$ the separation of slits is $d$ and the distance of screen is $D$ such that $D > > d > > \lambda$. If the fringe width is $\beta$, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is
A. $\dfrac{\beta }{4}$
B. $\dfrac{\beta }{3}$
C. $\dfrac{\beta }{6}$
D. $\dfrac{\beta }{2}$

Answer 1

Use the formulae for intensity of light at a point in Young’s double slits experiment, phase difference, path difference and fringe width. Using all these equations derive the relation for the phase difference between the two points, substitute this value of phase difference in the formula for intensity of light at a point and then rewrite this relation for intensity half of the maximum intensity of light.

Formulae used:
The intensity $I$ at a point in the Young’s double slit experiment is given by
$I = {I_{\max }}{\cos ^2}\left( {\dfrac{\phi }{2}} \right)$ …… (1)
Here, ${I_{\max }}$ is the maximum intensity and $\phi$ is the phase difference.
The phase difference $\phi$ in Young’s double slit experiment is given by
$\phi = \dfrac{{2\pi }}{\lambda }\Delta x$ …… (2)
Here, $\lambda$ is the wavelength of the light and $\Delta x$ is the path difference.
The path difference $\Delta x$ in Young’s double slit experiment is given by
$\Delta x = \dfrac{{yd}}{D}$ …… (3)
Here, $y$ is the distance between the two points of difference intensities on the screen, $d$ is the distance between the two slits and $D$ is the distance between the slit and screen.
The fringe width $\beta$ is given by
$\beta = \dfrac{{\lambda D}}{d}$ …… (4)
Here, $\lambda$ is the wavelength of the light, $D$ is the distance between the slits and screen and $d$ is the distance between the slits.

Complete step by step answer:
We have given that the distance between the two slits is $d$ and the distance between the slits and the screen is $D$. The wavelength of the light used in the Young’s double slit experiment is $\lambda$ and the fringe width is $\beta$.
Also, we have given that
$D > > d > > \lambda$

Let $y$ be the distance between the two points on the screen at which intensity of light is maximum and half of the maximum value.
Substitute $\dfrac{{yd}}{D}$ for $\Delta x$ in equation (2).
$\phi = \dfrac{{2\pi }}{\lambda }\dfrac{{yd}}{D}$
$\Rightarrow \phi = \dfrac{{2\pi }}{\lambda }\dfrac{{yd}}{D}$

Substitute $\dfrac{{\lambda D}}{d}$ for $\beta$ in the above equation.
$\Rightarrow \phi = \dfrac{{2\pi y}}{\beta }$
Substitute $\dfrac{{2\pi y}}{\beta }$ for $\phi$ in equation (1).
$I = {I_{\max }}{\cos ^2}\left( {\dfrac{{\dfrac{{2\pi y}}{\beta }}}{2}} \right)$
$\Rightarrow I = {I_{\max }}{\cos ^2}\left( {\dfrac{{\pi y}}{\beta }} \right)$

Let us consider the intensity at a point is $\dfrac{{{I_{\max }}}}{2}$ half of the maximum value. Substitute $\dfrac{{{I_{\max }}}}{2}$ for $I$ in the above equation.
$\Rightarrow \dfrac{{{I_{\max }}}}{2} = {I_{\max }}{\cos ^2}\left( {\dfrac{{\pi y}}{\beta }} \right)$
$\Rightarrow \dfrac{1}{2} = {\cos ^2}\left( {\dfrac{{\pi y}}{\beta }} \right)$
$\Rightarrow \cos \left( {\dfrac{{\pi y}}{\beta }} \right) = \dfrac{1}{{\sqrt 2 }}$
$\Rightarrow \dfrac{{\pi y}}{\beta } = \dfrac{\pi }{4}$
$\therefore y = \dfrac{\beta }{4}$
Therefore, the distance between the point of maximum intensity and intensity half of the maximum value is $\dfrac{\beta }{4}$.

Hence, the correct option is A.

Note: One can also solve the same question by another method. One can use the formula for resultant intensity of the light from two intensities of the light. Then calculate the value of the phase difference for intensity half of the maximum value. Calculate the value of path difference and determine the value of the distance between the points of maximum intensity and intensity half of the maximum value.