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MHT CET · Physics · Wave Optics

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. A \(\frac{\lambda\left(D_2-D_1\right)}{\left(W_1-w_2\right)}\)
  2. B \(\frac{\lambda\left(\mathrm{W}_1-\mathrm{W}_2\right)}{\left(\mathrm{D}_1-\mathrm{D}_2\right)}\)
  3. C \(\frac{\lambda\left(\mathrm{W}_2-\mathrm{W}_1\right)}{\left(\mathrm{D}_1-\mathrm{D}_2\right)}\)
  4. D \(\frac{\lambda\left(\mathrm{D}_1-\mathrm{D}_2\right)}{\left(\mathrm{W}_1-\mathrm{W}_2\right)}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{\lambda\left(\mathrm{D}_1-\mathrm{D}_2\right)}{\left(\mathrm{W}_1-\mathrm{W}_2\right)}\)

Step-by-step Solution

Detailed explanation

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