MHT CET · Physics · Wave Optics
In YDSE, the distance of the slits from the screen is increased by \(25 \%\) and the separation between the slits is halved. If ' \(\mathrm{W}\) ' represents the original fringe width, the new fringe width is
- A \(2 \mathrm{~W}\)
- B \(2.5 \mathrm{~W}\)
- C 4W
- D \(1.5 \mathrm{~W}\)
Answer & Solution
Correct Answer
(B) \(2.5 \mathrm{~W}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{E}=\frac{\lambda \mathrm{D}}{\mathrm{d}} \\ & \therefore \frac{\mathrm{W}_2}{\mathrm{~W}_1}=\frac{\mathrm{D}_2}{\mathrm{D}_1} \cdot \frac{\mathrm{d}_1}{\mathrm{~d}_2} \\ & \mathrm{D}_2=1.25 \mathrm{D}_1 \text { and } \mathrm{d}_2=\frac{\mathrm{d}_1}{2} \\ & \therefore \frac{\mathrm{W}_2}{\mathrm{~W}_1}=1.25 \times 2=2.5 \\ & \therefore \mathrm{W}_2=2.5 \mathrm{~W}\end{aligned}\)
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