Question

In an interference experiment, interference was observed using waves associated with electrons having de-Broglie wavelength \(\lambda\). The electrons are produced by using an electron-gun having the potential difference 'V'. To increase the fringe width in an interference pattern,

• A. the slit separation is increased.
• B. the screen is moved closer to interfering slits.
• C. electron gun voltage be increased.
• D. electron gun voltage be decreased.

Hint:

Fringe width is proportional to the wavelength, and the de Broglie wavelength of the electron grows when the accelerating voltage is lowered.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The fringe width is \(\beta = \dfrac{\lambda D}{d}\). It grows with a larger wavelength or a larger screen distance, and shrinks with a larger slit separation.

Step 2: Link to the voltage
Electrons accelerated through V have \(\lambda = \dfrac{h}{\sqrt{2meV}}\), so \(\lambda \propto \dfrac{1}{\sqrt{V}}\).

Step 3: Compare the options
Increasing the slit separation or moving the screen closer reduces \(\beta\). A higher gun voltage reduces \(\lambda\), which also reduces \(\beta\). Lowering the voltage increases \(\lambda\) and hence \(\beta\).

Final Answer:
The fringe width increases when the electron gun voltage is decreased. \[ \boxed{\text{(D) }\text{electron gun voltage is decreased}} \]