Question

A Young’s double-slit experiment is performed using monochromatic light of wavelength $\lambda$. The intensity of light at a point on the screen, where the path difference is $\lambda$, is $k$ units. The intensity of light at a point where the path difference is $\frac{\lambda }{6}$ is given by $\frac{nk}{12}$, where $n$ is an integer. The value of $n$ is __________.

Answer 1

Step 1. Given data

The intensity of the light when the path difference is $\lambda ,I=kunits$

The intensity of the light when the path difference is $\frac{\lambda }{6},I=\frac{nk}{12}$

Step 2. Finding the phase difference, $\varphi$ when the path difference is $\frac{\lambda }{6},$

$I=\frac{nk}{12}\dots \left(1\right)$

By using the formula of phase difference,

Phase difference $=\frac{2\mathrm{\pi }}{\lambda }\times$path difference

$\varphi =\frac{2\mathrm{\pi }}{\lambda }\times \frac{\lambda }{6}$

$\varphi =\frac{\mathrm{\pi }}{3}$

Step 3. Finding the value of $n$

By using the formula of intensity, $I$

$I=k{\cos }^2\frac{\varphi }{2}$ [$k$ is maximum intensity]

$I=k{\cos }^2\frac{2\mathrm{\pi }}{6}$

$I=\frac{3k}{4}\dots \left(2\right)$

Now, comparing equation $\left(1\right)$ and equation $\left(2\right)$ we can write,

$$\begin{gathered} \frac{nk}{12}=\frac{3k}{4} \\ n=\frac{3k}{4}\times \frac{12}{k} \end{gathered}$$

$n=9$

Therefore, the value of $n$ is $9$.