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 ' $\mathrm{d}$ ' is kept constant. If the distance between slit and eyepiece ' $D$ ' is varied as $D_1, D_2, D_3 \& D_4$ and corresponding measured fringe widths are $Z_1, Z_2, Z_3$ and $Z_4$ then
  1. $\mathrm{Z}_1 \mathrm{D}_1=\mathrm{Z}_2 \mathrm{D}_2=\mathrm{Z}_3 \mathrm{D}_3=\mathrm{Z}_4 \mathrm{D}_4$
  2. $\frac{Z_1}{D_1}=\frac{Z_2}{D_2}=\frac{Z_3}{D_3}=\frac{Z_4}{D_4}$
  3. $\mathrm{D}_1 \sqrt{\mathrm{Z}_1}=\mathrm{D}_2 \sqrt{\mathrm{Z}_2}=\mathrm{D}_3 \sqrt{\mathrm{Z}_3}=\mathrm{D}_4 \sqrt{\mathrm{Z}_4}$
  4. $\quad Z_1 \sqrt{D_1}=Z_2 \sqrt{D_2}=Z_3 \sqrt{D_3}=Z_4 \sqrt{D_4}$

Solution

Fringe width $Z=\frac{\lambda D}{d}$ $\therefore \quad \frac{Z}{D}=\frac{\lambda}{d}=$ constant, as $\lambda$ and $d$ are constant $\therefore \quad \frac{Z_1}{D_1}=\frac{Z_2}{D_2}=\frac{Z_3}{D_3}=\frac{Z_4}{D_4}$

Asked in: MHT CET 2023 (13 May Shift 1)

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