ExamBro
ExamBro
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
\(
\left(\cos 45^{\circ}=\frac{1}{\sqrt{2}}=\sin 45^{\circ}\right)
\)

  1. A \(2: 1\)
  2. B \(1: 4\)
  3. C \(1: 2\)
  4. D \(4: 1\)
Verified Solution

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, \(I=I_0 \cos ^2\left(\frac{\pi}{4}\right)=\frac{I_0}{2}\)