MHT CET · Physics · Wave Optics
In Young's double slit experiment, the wavelength of light used is ' \(\lambda\) '. The intensity at a point is ' \(I\) ' where path difference is \(\left(\frac{\lambda}{4}\right)\). If \(I_0\) denotes the maximum intensity, then the ratio \(\left(\frac{\mathrm{I}}{\mathrm{I}_0}\right)\) is
\(\left(\sin \frac{\pi}{4}=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right)\)
- A \(\frac{1}{\sqrt{2}}\)
- B \(\frac{1}{2}\)
- C \(\frac{3}{4}\)
- D \(\frac{\sqrt{3}}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{2}\)
Step-by-step Solution
Detailed explanation
Phase difference, \(\Delta \phi=\left(\frac{2 \pi}{\lambda}\right) \Delta l\)
For path difference \(\frac{\lambda}{4}\),
Phase difference \(\Delta \phi=\frac{\pi}{2}\)
Using, \(\mathrm{I}=\mathrm{I}_0 \cos ^2 \frac{\phi}{2}\)
\(\begin{array}{ll}
\therefore & \frac{\mathrm{I}}{\mathrm{I}_0}=\cos ^2 \frac{\phi}{2}=\cos ^2\left(\frac{\pi}{4}\right) \\
\therefore & \frac{\mathrm{I}}{\mathrm{I}_0}=\frac{1}{2}
\end{array}\)
For path difference \(\frac{\lambda}{4}\),
Phase difference \(\Delta \phi=\frac{\pi}{2}\)
Using, \(\mathrm{I}=\mathrm{I}_0 \cos ^2 \frac{\phi}{2}\)
\(\begin{array}{ll}
\therefore & \frac{\mathrm{I}}{\mathrm{I}_0}=\cos ^2 \frac{\phi}{2}=\cos ^2\left(\frac{\pi}{4}\right) \\
\therefore & \frac{\mathrm{I}}{\mathrm{I}_0}=\frac{1}{2}
\end{array}\)
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