Question

In Young's double slit experiment, the separation between the slits is decreased to \(\left(\frac{1}{3}\right)^{th}\) of its original distance and the distance between the slits and the screen is doubled. The fringe width

• A. remains unchanged.
• B. becomes \(2\) times the original.
• C. becomes \(3\) times the original.
• D. becomes \(6\) times the original.

Hint:

Fringe width is proportional to the screen distance and inversely proportional to the slit separation.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
In Young's double slit experiment the fringe width is \(\beta=\dfrac{\lambda D}{d}\).

Step 2: Write the new values
The new slit separation is \(d'=\dfrac{d}{3}\) and the new screen distance is \(D'=2D\).

Step 3: Calculate the new fringe width
\(\beta'=\dfrac{\lambda D'}{d'}=\dfrac{\lambda(2D)}{d/3}=6\,\dfrac{\lambda D}{d}=6\beta\)

Step 4: Compare the options
The fringe width becomes \(6\) times the original, option (D).

Final Answer:
The fringe width becomes 6 times the original. \[ \boxed{\text{(D) }\text{becomes 6 times the original}} \]