Q79 / 90 · KVPY 2015 (SB/SX) · Physics
A beam of monoenergetic electrons, which have been accelerated from rest by a potential $\mathrm{U}$, is used to form an interference pattern in a Young's Double Slit experiment. The electrons are now accelerated by potential $4 \mathrm{U}$. Then the fringe width
- Aremains the same
- Correct answerBis half the original fringe width
- Cis twice the original fringe width
- Dis one-fourth the original fringe width
Solution
$\mathrm{eV}=\frac{1}{2} \mathrm{mv}^{2} \mathrm{so}, \mathrm{eV}=\frac{p^{2}}{2 \mathrm{~m}}$
and
$$
\lambda=\frac{h}{p}, \Rightarrow V \propto \frac{1}{\lambda^{2}} \text { and } \beta=\frac{n \lambda D}{d}
$$
So as potential U becomes $4 \mathrm{U}$ then $\beta$ becomes half
Hence, the correct answer is option: 'B'.