Question

In Young's double slit experiment, the slits are separated by $0.1mm$ and they are $0.5m$ from the screen. The wave length used is $500nm$. Find the distance between $7^{th}$ maxima and ${11}^{th}$ minima on the screen.

Answer 1

Given,

Slits are seperated by $d=0.1mm$

Distance from the screen is $D=0.5m$

So, the position of the seventh maxima is $7\lambda$

And the position of the eleventh minima is $\frac{21}{2}\lambda$

So, the diffrence between seventh maximaand eleventh minima is:

$\frac{21}{2}-7=\frac{7}{2}$

So, the doistance, between them is:

$\Delta y=\frac{\Delta xD}{d}$

Where $\Delta y$ is the difference between them,

$\Delta x$ ias the difference,

Then,

$=\frac{0.5\times 7\times 500\times {10}^{-9}}{2\times 0.1\times {10}^{-3}}$

$=8.75\times {10}^{-3}$