MHT CET · Physics · Wave Optics
Young's double slit experiment, the intensity at a point where the path difference is \(\frac{{ }^{\prime}}{4}\) ' is ' \(I\) '. If the maximum intensity is \(\mathrm{I}_0\) then the ration \(\frac{\mathrm{I}_0}{\mathrm{I}}\) is
\(
\left(\cos 45^{\circ}=\frac{1}{\sqrt{2}}=\sin 45^{\circ}\right)
\)
- A \(2: 1\)
- B \(1: 4\)
- C \(1: 2\)
- D \(4: 1\)
Answer & Solution
Correct Answer
(A) \(2: 1\)
Step-by-step Solution
Detailed explanation
For a path difference of \(f\), the intensity at point is,
\(I=I_0 \cos ^2 \frac{\phi}{2}\)
where, \(\mathrm{I}_0\) is the maximum intensity
Here,
\(\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2}\)
Thus, \(I=I_0 \cos ^2\left(\frac{\pi}{4}\right)=\frac{I_0}{2}\)
\(I=I_0 \cos ^2 \frac{\phi}{2}\)
where, \(\mathrm{I}_0\) is the maximum intensity
Here,
\(\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2}\)
Thus, \(I=I_0 \cos ^2\left(\frac{\pi}{4}\right)=\frac{I_0}{2}\)
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