MHT CET · Physics · Wave Optics
In Young's experiment, fringes are obtained on a screen placed a distance \(75 \mathrm{~cm}\) from the slits. When the separation between two narrow slits is doubled, then the fringe width is decrease. In order to obtain the initial fringe width, the screen should be moved through.
- A 150 cm away from the slits
- B 75 cm towards the slits
- C 75 cm away from the slits
- D 150 cm towards the slits
Answer & Solution
Correct Answer
(C) 75 cm away from the slits
Step-by-step Solution
Detailed explanation
Fringe width, \(\beta=\frac{\lambda D_1}{d_1}=\frac{\lambda D_2}{d_2}\)
\(
\begin{aligned}
& \therefore \frac{d_2}{d_1}=\frac{D_2}{D_1} \\
& \therefore 2=\frac{D_2}{D_1} \\
& \therefore D_2=2 D_1=2 \times 75=150 \mathrm{~cm}
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \frac{d_2}{d_1}=\frac{D_2}{D_1} \\
& \therefore 2=\frac{D_2}{D_1} \\
& \therefore D_2=2 D_1=2 \times 75=150 \mathrm{~cm}
\end{aligned}
\)
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