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Question: Answered & Verified by Expert
In a biprism experiment, monochromatic light of wavelength $(\lambda)$ is used. The distance between two coherent sources is kept constant. If the distance between slit and eyepiece (D) is varied as $D_{1}, D_{2}, D_{3}$ and $D_{4}$, the corresponding measured fringe widths are $\mathrm{z}_{1}, \mathrm{z}_{2}, \mathrm{z}_{3}$ and $\mathrm{z}_{4}$ then
PhysicsWave OpticsMHT CETMHT CET 2020 (12 Oct Shift 2)
Options:
  • A $\frac{z_{1}}{D_{1}}=\frac{z_{2}}{D_{2}}=\frac{z_{3}}{D_{3}}=\frac{z_{4}}{D_{4}}$
  • B $\mathrm{z}_{1} \mathrm{D}_{1}=\mathrm{z}_{2} \mathrm{D}_{2}=\mathrm{z}_{3} \mathrm{D}_{3}=\mathrm{z}_{4} \mathrm{D}_{4}$
  • C $\mathrm{z}_{1} \sqrt{\mathrm{D}_{1}}=\mathrm{z}_{2} \sqrt{\mathrm{D}_{2}}=\mathrm{z}_{3} \sqrt{\mathrm{D}_{3}}=\mathrm{z}_{4} \sqrt{\mathrm{D}_{4}}$
  • D $\mathrm{z}_{1} \mathrm{D}_{1}^{2}=\mathrm{z}_{2} \mathrm{D}_{2}^{2}=\mathrm{z}_{3} \mathrm{D}_{3}^{2}=\mathrm{z}_{4} \mathrm{D}_{4}^{2}$
Solution:
1171 Upvotes Verified Answer
The correct answer is: $\frac{z_{1}}{D_{1}}=\frac{z_{2}}{D_{2}}=\frac{z_{3}}{D_{3}}=\frac{z_{4}}{D_{4}}$
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