Answer to Question #119591 in Optics for Ananya Sharma

Question

In a Young’s double slit experiment, the separation of eight bright fringes is 5mm when the wavelength used is 6200Å. The distance from the slits to the screen is 0.8m. Calculate the separation of the two slit.

Answer 1

Given, Position of $8^{th}$ bright fringes is $y=5mm=5\times 10^{-3}m$ .

Wavelength is $\lambda=6200A^{\circ}=62\times10^{-8}m$ ,Distance between slit and screen is $D=0.8m$ ,

Let $d$ is the distance between slits.

Clearly,

$$\tan(\theta)=\frac{y}{D}<<1\implies \tan(\theta)\approx\sin(\theta)\approx\theta=\frac{y}{D}$$

Thus,formula for position of bright fringes is

$$d\sin(\theta)=n\lambda\implies d=\frac{n\lambda D}{y}$$

Hence,

$$d=\frac{8\times 62\times 10^{-8}\times 0.8}{5\times 10^{-3}}\\ \implies d=79.36\times 10^{-5}m=793.6\mu m$$