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MHT CET · Physics · Wave Optics

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

  1. A \(\frac{z_{1}}{D_{1}}=\frac{z_{2}}{D_{2}}=\frac{z_{3}}{D_{3}}=\frac{z_{4}}{D_{4}}\)
  2. B \(\mathrm{z}_{1} \mathrm{D}_{1}=\mathrm{z}_{2} \mathrm{D}_{2}=\mathrm{z}_{3} \mathrm{D}_{3}=\mathrm{z}_{4} \mathrm{D}_{4}\)
  3. C \(\mathrm{z}_{1} \sqrt{\mathrm{D}_{1}}=\mathrm{z}_{2} \sqrt{\mathrm{D}_{2}}=\mathrm{z}_{3} \sqrt{\mathrm{D}_{3}}=\mathrm{z}_{4} \sqrt{\mathrm{D}_{4}}\)
  4. D \(\mathrm{z}_{1} \mathrm{D}_{1}^{2}=\mathrm{z}_{2} \mathrm{D}_{2}^{2}=\mathrm{z}_{3} \mathrm{D}_{3}^{2}=\mathrm{z}_{4} \mathrm{D}_{4}^{2}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{z_{1}}{D_{1}}=\frac{z_{2}}{D_{2}}=\frac{z_{3}}{D_{3}}=\frac{z_{4}}{D_{4}}\)

Step-by-step Solution

Detailed explanation

\(z = \frac{\lambda D}{d}\) Since \(\lambda\) and \(d\) are constant, \(z \propto D\).
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