Question

In a Young’s double slit experiment, light of $500nm$ is used to produce an interference pattern. When the distance between the slits is$0.05mm$, the angular width (in degree) of the fringes formed on the distance screen is close to:

• A. $0.17°$
• B. $0.07°$
• C. $0.57°$
• D. $1.7°$

Answer 1

The correct option is C

$0.57°$

Step 1: Given

Wavelength, $$\begin{gathered} \lambda =500nm \\ =500\times {10}^{-9}m \end{gathered}$$

Distance between the slits, $$\begin{gathered} d=0.05mm \\ =5\times {10}^{-5}m \end{gathered}$$

Step 2: Formula used

In Young's double-slit experiment, the angular width is given by $\theta =\frac{\lambda }{d}$.

Step 3: Determine the angular width

In Young's double-slit experiment, the angular width is given by

$$\begin{gathered} \theta =\frac{\lambda }{d} \\ =\frac{500\times {10}^{-9}}{5\times {10}^{-5}} \\ =0.01rad \end{gathered}$$

To convert radian into degree:

$$\begin{gathered} =0.01\times \frac{180}{\mathrm{\pi }} \\ =0.57° \end{gathered}$$

Therefore, option C is the correct option.