MHT CET · Physics · Wave Optics
In Young's double slit experiment using monochromatic light of wavelength ' \(\lambda\) ', the intensity of light at a point on the screen where path difference ' \(\lambda\) ' is K units. The intensity of light at a point where the path difference is \(\frac{\lambda}{6}\) is \(\left[\cos \frac{\pi}{6}=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\right]\)
- A K
- B \(\frac{3 \mathrm{~K}}{4}\)
- C \(\frac{\mathrm{K}}{2}\)
- D \(\frac{\mathrm{K}}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{3 \mathrm{~K}}{4}\)
Step-by-step Solution
Detailed explanation
The intensity is given by
\(I=4 I_0 \cos ^2 \frac{\phi}{2}\)
Maximum intensity \(\mathrm{K}=4 \mathrm{I}_0\) when \(\phi=0\) when path difference is \(\frac{\lambda}{6}\),
\(\begin{aligned}
& \phi=\frac{2 \pi}{\lambda} \times \text { path difference }=\frac{\pi}{3} \\
\therefore \quad & I=K \cos ^2\left(\frac{\pi}{6}\right)=K\left(\frac{1}{4}\right)=\frac{3 K}{4}
\end{aligned}\)
...[From(i) and (ii)]
\(I=4 I_0 \cos ^2 \frac{\phi}{2}\)
Maximum intensity \(\mathrm{K}=4 \mathrm{I}_0\) when \(\phi=0\) when path difference is \(\frac{\lambda}{6}\),
\(\begin{aligned}
& \phi=\frac{2 \pi}{\lambda} \times \text { path difference }=\frac{\pi}{3} \\
\therefore \quad & I=K \cos ^2\left(\frac{\pi}{6}\right)=K\left(\frac{1}{4}\right)=\frac{3 K}{4}
\end{aligned}\)
...[From(i) and (ii)]
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