MHT CET · Physics · Wave Optics
In Young's double slit experiment the separation between the slits is doubled without changing other setting of the experiment to obtain same fringe width, the distance ' \(\mathrm{D}\) ' of the screen from slit should be made
- A \(\frac{D}{2}\)
- B \(\frac{\mathrm{D}}{\sqrt{2}}\)
- C \(2 \mathrm{D}\)
- D \(4 \mathrm{D}\)
Answer & Solution
Correct Answer
(C) \(2 \mathrm{D}\)
Step-by-step Solution
Detailed explanation
Fringe width,
\(\mathrm{W}=\frac{\lambda \mathrm{D}}{\mathrm{d}}\)
Given: \(\mathrm{d}=2 \mathrm{~d}\)
\(\Rightarrow \mathrm{W}^{\prime}=\frac{\lambda \mathrm{D}^{\prime}}{2 \mathrm{~d}}\)
Since given,
\(\begin{aligned}
& \mathrm{W}=\mathrm{W}^{\prime} \\
\therefore & \frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{\lambda \mathrm{D}^{\prime}}{2 \mathrm{~d}} \\
\therefore & \mathrm{D}^{\prime}=2 \mathrm{D}
\end{aligned}\)
\(\mathrm{W}=\frac{\lambda \mathrm{D}}{\mathrm{d}}\)
Given: \(\mathrm{d}=2 \mathrm{~d}\)
\(\Rightarrow \mathrm{W}^{\prime}=\frac{\lambda \mathrm{D}^{\prime}}{2 \mathrm{~d}}\)
Since given,
\(\begin{aligned}
& \mathrm{W}=\mathrm{W}^{\prime} \\
\therefore & \frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{\lambda \mathrm{D}^{\prime}}{2 \mathrm{~d}} \\
\therefore & \mathrm{D}^{\prime}=2 \mathrm{D}
\end{aligned}\)
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