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

In Young's double slit experiment, intensity at a point is given by

$I=I_0{\cos }^2\frac{\varphi }{2}......\left(i\right)$

where, $\varphi =$ phase difference,

Using phase difference, $\varphi =\frac{2\pi }{\lambda }\times \text{path difference}$

For path difference $\lambda ,$ phase difference ${\varphi }_1=2\pi$

For path difference, $\frac{\lambda }{6},$ phase difference ${\varphi }_2=\frac{\pi }{3}$

Using equation (i),

$\frac{I_1}{I_2}=\frac{{\cos }^2\left(\frac{{\varphi }_1}{2}\right)}{{\cos }^2\left(\frac{{\varphi }_2}{2}\right)}=\frac{{\cos }^2\frac{2\pi }{2}}{{\cos }^2\left(\frac{\pi }{6}\right)}$

$\Rightarrow \frac{K}{I_2}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\Rightarrow I_2=\frac{3K}{4}=\frac{9K}{12}$

$\therefore n=9$