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MHT CET · Physics · Wave Optics

In Young's double slit experiment, the intensity at a point where path difference is \(\frac{\lambda}{6}\) ( \(\lambda\) being the wavelength of light used) is I'. If ' \(I_0\) ' denotes the maximum intensity, then \(\frac{I}{I_0}\) is equal to

  1. A \(\frac{\sqrt{3}}{2}\)
  2. B \(\frac{3}{4}\)
  3. C \(\frac{1}{\sqrt{2}}\)
  4. D \(\frac{1}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{3}{4}\)

Step-by-step Solution

Detailed explanation

If \(\mathrm{I}\) ' is the intensity of each wave, then the resultant intensity at a point is given by
\(
\mathrm{I}=4 \mathrm{I}^{\prime} \cos ^2 \frac{\phi}{2}
\)
For maximum intensity, phase difference \(\phi=0\)
\(
\therefore \mathrm{I}_0=4 \mathrm{I}^{\prime} \cos ^2 0=4 \mathrm{I}^{\prime}
\)
At a point where path difference is \(\frac{\lambda}{6}\), hence phase difference is
\(
\begin{aligned}
& \frac{\pi}{3} . \\
& \therefore \mathrm{I}=4 \mathrm{I}^{\prime} \cos ^2 \frac{\pi}{6}=3 \mathrm{I}^{\prime} \\
& \therefore \frac{\mathrm{I}}{\mathrm{I}_0}=\frac{3}{4}
\end{aligned}
\)
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