KCET · Physics · Wave Optics
The wavelength of the light used in Young's double slit experiment is \(\lambda\). The intensity at a point on the screen is I, where the path difference is \(\frac{\lambda}{6}\). If \(\mathrm{I}_{0}\) denotes the maximum intensity, then the ratio of I and \(\mathrm{I}_{0}\) is
- A \(0.866\)
- B \(0.5\)
- C \(0.707\)
- D \(0.75\)
Answer & Solution
Correct Answer
(D) \(0.75\)
Step-by-step Solution
Detailed explanation
Phase difference, \(\phi=\frac{2 \pi}{\lambda} \times\) path difference
\[
\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}=\frac{\pi}{3}=60^{\circ}
\]
Intensity, \(\mathrm{I}=\mathrm{I}_{0} \cos ^{2}\left(\frac{\phi}{2}\right)\)
\[
\frac{\mathrm{I}}{\mathrm{I}_{0}}=\cos ^{2}\left(30^{\circ}\right)=\left(\frac{\sqrt{3}}{2}\right)^{2}=0.75
\]
\[
\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}=\frac{\pi}{3}=60^{\circ}
\]
Intensity, \(\mathrm{I}=\mathrm{I}_{0} \cos ^{2}\left(\frac{\phi}{2}\right)\)
\[
\frac{\mathrm{I}}{\mathrm{I}_{0}}=\cos ^{2}\left(30^{\circ}\right)=\left(\frac{\sqrt{3}}{2}\right)^{2}=0.75
\]
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