TS EAMCET · Physics · Wave Optics
In Young's double slit experiment, intensity of light at a point on the screen where the path difference becomes \(\lambda\) is \(I\). The intensity at a point on the screen where the path difference becomes \(\frac{\lambda}{3}\) is,
- A \(\frac{I}{4}\)
- B \(\frac{I}{3}\)
- C \(\frac{2 I}{3}\)
- D \(3 I\)
Answer & Solution
Correct Answer
(A) \(\frac{I}{4}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \quad \text { In YDSE, } \mathrm{I}_1=\mathrm{I}, \Delta \mathrm{x}_1=\lambda, \Delta \mathrm{x}_2=\frac{\lambda}{3} \\ & \therefore \quad \phi_1=\frac{2 \pi}{\lambda} \Delta \mathrm{x}_1=\frac{2 \pi}{\lambda} \cdot \lambda=2 \pi \\ & \phi_2=\frac{2…
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