In a biprism experiment, monochromatic light of wavelength ' $\gamma$ ' is used. The distance between the…

In a biprism experiment, monochromatic light of wavelength ' $\gamma$ ' is used. The distance between the two coherent sources ' $d$ ' is kept constant. If the distance between slit and eyepiece ' $D$ ' is varied as $\mathrm{D}_1, \mathrm{D}_2, \mathrm{D}_3, \mathrm{D}_4$ and corresponding measured fringe widths are $\mathrm{W}_1, \mathrm{~W}_2, \mathrm{~W}_3, \mathrm{~W}_4$ then
  1. $\mathrm{W}_1 \mathrm{D}_1=\mathrm{W}_2 \mathrm{D}_2=\mathrm{W}_3 \mathrm{D}_3=\mathrm{W}_4 \mathrm{D}_4$
  2. $\frac{\mathrm{W}_1}{\mathrm{D}_1}=\frac{\mathrm{W}_2}{\mathrm{D}_2}=\frac{\mathrm{W}_3}{\mathrm{D}_3}=\frac{\mathrm{W}_4}{\mathrm{D}_4}$
  3. $\mathrm{W}_1 \sqrt{\mathrm{D}_1}=\mathrm{W}_2 \sqrt{\mathrm{D}_2}=\mathrm{W}_3 \sqrt{\mathrm{D}_3}=\mathrm{W}_3 \sqrt{\mathrm{D}_3}$
  4. $\mathrm{D}_1 \sqrt{\mathrm{~W}_1}=\mathrm{D}_2 \sqrt{\mathrm{~W}_2}=\mathrm{D}_3 \sqrt{\mathrm{~W}_3}=\mathrm{D}_4 \sqrt{\mathrm{~W}_4}$

Solution

$\begin{array}{ll} & \text { Fringe width } \mathrm{W}=\frac{\lambda \mathrm{D}}{\mathrm{d}} \\ \therefore \quad & \frac{\mathrm{W}}{\mathrm{D}}=\frac{\lambda}{\mathrm{d}} \\ & \frac{\mathrm{W}}{\mathrm{D}}=\text { constant, as } \lambda \text { and d are constant } \\ \therefore \quad & \frac{\mathrm{W}_1}{\mathrm{D}_1}=\frac{\mathrm{W}_2}{\mathrm{D}_2}=\frac{\mathrm{W}_3}{\mathrm{D}_3}=\frac{\mathrm{W}_4}{\mathrm{D}_4}\end{array}$

Asked in: MHT CET 2024 (04 May Shift 2)

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