MHT CET · Physics · Wave Optics

In Young's double slit experiment using monochromatic light of wavelength ' \(\lambda\) ', the maximum intensity of light at a point on the screen is ' \(K\) ' units. The intensity of light at a point where the path difference is \(\frac{\lambda}{6}\), is \(\left(\cos 60^{\circ}=\sin 30^{\circ}=0.5, \sin 60^{\circ}=\cos 30^{\circ}=\sqrt{3} / 2\right)\).

A \(\frac{3 \mathrm{~K}}{4}\)
B \(\frac{\mathrm{K}}{4}\)
C \(\frac{\mathrm{K}}{2}\)
D K

Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{3 \mathrm{~K}}{4}\)

Step-by-step Solution

Detailed explanation

The intensity is given by
\(\mathrm{I}=4 \mathrm{I}_0 \cos ^2 \frac{\phi}{2}...(i)\)
Maximum intensity \(\mathrm{K}=4 \mathrm{I}_0\) when \(\phi=0\)
when path difference is \(\frac{\lambda}{6}\),
\(\begin{aligned}
& \phi=\frac{2 \pi}{\lambda} \times \text { path difference }=\frac{\pi}{3} ...(ii)\\
\therefore \quad & I=K \cos ^2\left(\frac{\pi}{6}\right)=K\left(\frac{3}{4}\right)=\frac{3 K}{4}
\end{aligned}\)
...[From(i) and (ii)]

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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