MHT CET · Physics · Wave Optics
In Young's double slit experiment the intensity at a point on the screen is \(K\), where path difference is \(\lambda\). What will be the intensity at the point where path difference is \(\frac{\lambda}{4}\) ?
- A \(\frac{K}{4}\)
- B \(\frac{K}{2}\)
- C \(K\)
- D 0
Answer & Solution
Correct Answer
(B) \(\frac{K}{2}\)
Step-by-step Solution
Detailed explanation
\(I=2 I_0(1+\cos \phi)\)
\(\phi=\frac{2 \pi x}{\lambda}\), where \(x\) is the path as difference.
\(\therefore K=2 I_0\left(1+\cos \left(\frac{2 \pi}{\lambda} \times \lambda\right)\right)=4 I_0\)
Now, for path difference \(\frac{\lambda}{4}\)
\(\begin{aligned}
& K^{\prime}=2 I_0\left(1+\cos \left(\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}\right)\right)=2 I_0 \\
& \therefore K^{\prime}=\frac{K}{2}
\end{aligned}\)
\(\phi=\frac{2 \pi x}{\lambda}\), where \(x\) is the path as difference.
\(\therefore K=2 I_0\left(1+\cos \left(\frac{2 \pi}{\lambda} \times \lambda\right)\right)=4 I_0\)
Now, for path difference \(\frac{\lambda}{4}\)
\(\begin{aligned}
& K^{\prime}=2 I_0\left(1+\cos \left(\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}\right)\right)=2 I_0 \\
& \therefore K^{\prime}=\frac{K}{2}
\end{aligned}\)
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