Q118 / 150 · MHT CET 2020 (13 Oct Shift 2) · Physics
distance between slit and eyepiece. When the distance between two virtual
sources is changed from $\mathrm{d}_{\mathrm{A}}$ to $\mathrm{d}_{\mathrm{B}}$, then the fringe width is changed from $\mathrm{Z}_{\mathrm{A}}$ to
$\mathrm{Z}_{\mathrm{B}}$. The ratio $\mathrm{Z}_{\mathrm{A}}$ to $\mathrm{Z}_{\mathrm{B}}$ is
- A$\left(\frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{d}_{\mathrm{B}}}\right)^{2}$
- B$\left(\frac{\mathrm{d}_{\mathrm{A}}}{\mathrm{d}_{\mathrm{B}}}\right)$
- Correct answerC$\left(\frac{\mathrm{d}_{\mathrm{B}}}{\mathrm{d}_{\mathrm{A}}}\right)$
- D$\sqrt{\frac{\mathrm{d}_{\mathrm{B}}}{\mathrm{d}_{\mathrm{A}}}}$
Solution
Fringe width is given by $z=\frac{\lambda D}{d}$
$\therefore \frac{\mathrm{z}_{\mathrm{A}}}{\mathrm{Z}_{\mathrm{B}}}=\frac{\mathrm{d}_{\mathrm{B}}}{\mathrm{d}_{\mathrm{A}}}$
Hence, the correct answer is option: 'C'.