Question

The maximum intensity in Young's double-slit experiment is $I_0$. Distance between the slits is $d=5\lambda$, where $\lambda$ is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance $D=10d$ ?

• A. $\frac{I_0}{2}$
• B. $\frac{3I_0}{4}$
• C. $2I_0$
• D. $\frac{I_0}{4}$

Answer 1

The correct option is A $\frac{I_0}{2}$

Given, $d=5\lambda$, $D=10d$

Let the intensity of each source is $I$.
The resultant intensity at a point is, $I_R=4I{\cos }^2\left(\frac{\varphi }{2}\right)$
The maximum intensity is, $I_0=4I$

Now, the path difference is given by,
$\Delta x=\frac{yd}{D}$
For a point in front of one of the slits, $y=\frac{d}{2}=\frac{5\lambda }{2}$,
And, $D=10d=50\lambda$ $(\because d=5\lambda$)

So,
$\Delta x=\left(\frac{5\lambda }{2}\right)\left(\frac{5\lambda }{50\lambda }\right)=\frac{\lambda }{4}$

Corresponding phase difference will be,
$\Delta \varphi =\frac{2\pi }{\lambda }\times \left(\Delta x\right)=\left(\frac{2\pi }{\lambda }\right)\left(\frac{\lambda }{4}\right)=\frac{\pi }{2}$

$\therefore I_R=4I{\cos }^2\left(\frac{\Delta \varphi }{2}\right)$
$I_R=4I{\cos }^2\left(\frac{\pi }{4}\right)=2I=\frac{I_0}{2}$

Hence, $\left(A\right)$ is the correct answer.

$\begin{array}{c}\text{Why this question?}\\ \text{Note : If both slits have the same intensity of light i.e. I, and have zero phase difference}\\ \text{on the screen when interfering, then the maximum intensity will be 4I.}\end{array}$