Question

In Young's double-slit experiment two slits are separated by $2mm$ and the screen is placed one meter away. When the light of wavelength $500nm$ is used, the fringe separation will be:

• A. $0.75mm$
• B. $0.50mm$
• C. $1mm$
• D. $0.25mm$

Answer 1

The correct option is D

$0.25mm$

Step 1: Given information

The screen is placed, $D=1m$

Two slits are separated, $d=2\times {10}^{-3}m$

The light of wavelength, $\lambda =500\times {10}^{-9}m$

Step 2: Calculating the fringe width

Fringe width, $\beta =\frac{D\lambda }{d}$

Where $\beta =$fringe width, $D$$=$distance between the screen and the slit, $d=$distance between two slits, and $\lambda =$wavelength of light.

According to the question,

$$\begin{gathered} \Rightarrow \beta =\frac{1\times 500\times {10}^{-9}}{2\times {10}^{-3}} \\ \Rightarrow \beta =0.25mm \end{gathered}$$

Hence the correct option is (D).