In a biprism experiment, monochromatic light of wavelength $(\lambda)$ is used. The distance between two…

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. $\frac{z_{1}}{D_{1}}=\frac{z_{2}}{D_{2}}=\frac{z_{3}}{D_{3}}=\frac{z_{4}}{D_{4}}$
  2. $\mathrm{z}_{1} \mathrm{D}_{1}=\mathrm{z}_{2} \mathrm{D}_{2}=\mathrm{z}_{3} \mathrm{D}_{3}=\mathrm{z}_{4} \mathrm{D}_{4}$
  3. $\mathrm{z}_{1} \sqrt{\mathrm{D}_{1}}=\mathrm{z}_{2} \sqrt{\mathrm{D}_{2}}=\mathrm{z}_{3} \sqrt{\mathrm{D}_{3}}=\mathrm{z}_{4} \sqrt{\mathrm{D}_{4}}$
  4. $\mathrm{z}_{1} \mathrm{D}_{1}^{2}=\mathrm{z}_{2} \mathrm{D}_{2}^{2}=\mathrm{z}_{3} \mathrm{D}_{3}^{2}=\mathrm{z}_{4} \mathrm{D}_{4}^{2}$

Solution

- $\lambda$ is the wavelength of light, - $D$ is the distance between the slit and the eyepiece, - $d$ is the separation between the two coherent sources. When $D$ is varied, the fringe width $\beta$ changes proportionally. Given that the distances $D_1, D_2, \ldots$ and corresponding fringe widths $z_1, z_2, \ldots$ are measured, the relationship is: $z_1 D_1=z_2 D_2=z_3 D_3=z_4 D_4$ From the given options, Answer (1) correctly represents this proportional relationship: $z_1 D_1=z_2 D_2=z_3 D_3=z_4 D_4$

Asked in: MHT CET 2020 (12 Oct Shift 2)

Practice more Wave Optics questions on Aicharya