In biprism experiment, the wavelength $\lambda$ is used to obtain an interference pattern. Then fringe width…

In biprism experiment, the wavelength $\lambda$ is used to obtain an interference pattern. Then fringe width is $\mathrm{W}_1$ at distance $\mathrm{D}_1$ from the screen when the screen is moved towards biprism, fringe width becomes $\mathrm{W}_2$ at distance $\mathrm{D}_2$ The distance between two virtual images of the slit is
  1. $\frac{\lambda\left(D_2-D_1\right)}{\left(W_1-w_2\right)}$
  2. $\frac{\lambda\left(\mathrm{W}_1-\mathrm{W}_2\right)}{\left(\mathrm{D}_1-\mathrm{D}_2\right)}$
  3. $\frac{\lambda\left(\mathrm{W}_2-\mathrm{W}_1\right)}{\left(\mathrm{D}_1-\mathrm{D}_2\right)}$
  4. $\frac{\lambda\left(\mathrm{D}_1-\mathrm{D}_2\right)}{\left(\mathrm{W}_1-\mathrm{W}_2\right)}$

Solution

Fringe width $=\frac{\lambda D}{d}$ $\mathrm{W}_1=\frac{\lambda \mathrm{D}_1}{\mathrm{~d}}$ $\mathrm{w}_2=\frac{\lambda \mathrm{D}_2}{\mathrm{~d}}$ $\begin{aligned} & \Delta \mathrm{w}=\mathrm{w}_1-\mathrm{w}_2 \\ & =\frac{\lambda}{\mathrm{d}}\left(\mathrm{D}_1-\mathrm{D}_2\right) \\ & \Rightarrow \mathrm{d}=\frac{\lambda\left(\mathrm{D}_1-\mathrm{D}_2\right)}{\left(\mathrm{w}_1-\mathrm{w}_2\right)}\end{aligned}$

Asked in: MHT CET 2022 (07 Aug Shift 2)

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