MHT CET · Physics · Wave Optics
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
- A \(\mathrm{W}_1 \mathrm{D}_1=\mathrm{W}_2 \mathrm{D}_2=\mathrm{W}_3 \mathrm{D}_3=\mathrm{W}_4 \mathrm{D}_4\)
- B \(\frac{\mathrm{W}_1}{\mathrm{D}_1}=\frac{\mathrm{W}_2}{\mathrm{D}_2}=\frac{\mathrm{W}_3}{\mathrm{D}_3}=\frac{\mathrm{W}_4}{\mathrm{D}_4}\)
- C \(\mathrm{W}_1 \sqrt{\mathrm{D}_1}=\mathrm{W}_2 \sqrt{\mathrm{D}_2}=\mathrm{W}_3 \sqrt{\mathrm{D}_3}=\mathrm{W}_3 \sqrt{\mathrm{D}_3}\)
- D \(\mathrm{D}_1 \sqrt{\mathrm{~W}_1}=\mathrm{D}_2 \sqrt{\mathrm{~W}_2}=\mathrm{D}_3 \sqrt{\mathrm{~W}_3}=\mathrm{D}_4 \sqrt{\mathrm{~W}_4}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{W}_1}{\mathrm{D}_1}=\frac{\mathrm{W}_2}{\mathrm{D}_2}=\frac{\mathrm{W}_3}{\mathrm{D}_3}=\frac{\mathrm{W}_4}{\mathrm{D}_4}\)
Step-by-step Solution
Detailed explanation
\(\text {Fringe width } \mathrm{W}=\frac{\lambda \mathrm{D}}{\mathrm{d}} \)
\( \therefore \frac{\mathrm{W}}{\mathrm{D}}=\frac{\lambda}{\mathrm{d}} \)
\( \frac{\mathrm{W}}{\mathrm{D}}=\text {constant, as } \lambda \text {and d are constant } \)
\( \therefore \frac{\mathrm{W}_1}{\mathrm{D}_1}=\frac{\mathrm{W}_2}{\mathrm{D}_2}=\frac{\mathrm{W}_3}{\mathrm{D}_3}=\frac{\mathrm{W}_4}{\mathrm{D}_4}\)
\( \therefore \frac{\mathrm{W}}{\mathrm{D}}=\frac{\lambda}{\mathrm{d}} \)
\( \frac{\mathrm{W}}{\mathrm{D}}=\text {constant, as } \lambda \text {and d are constant } \)
\( \therefore \frac{\mathrm{W}_1}{\mathrm{D}_1}=\frac{\mathrm{W}_2}{\mathrm{D}_2}=\frac{\mathrm{W}_3}{\mathrm{D}_3}=\frac{\mathrm{W}_4}{\mathrm{D}_4}\)
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