TS EAMCET · Physics · Wave Optics
In the Young's double slit experiment, the resultant intensity at a point on the screen is \(75 \%\) of the maximum intensity of the bright fringe. Then the phase difference between the two interfering rays at that point is
- A \(\frac{\pi}{6}\)
- B \(\frac{\pi}{4}\)
- C \(\frac{\pi}{3}\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{3}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Given, } I_R=75 \% \text { of } I_{\max } \\ &=\frac{3}{4} I_{\max } \\ &=\frac{3}{4}\left(4 a^2\right)=3 a^2 \\ & \Rightarrow \quad 4 a^2 \cos ^2 \frac{\phi}{2}=3 a^2 \\ & \cos ^2 \frac{\phi}{2}=\frac{3}{4} \text { or } \cos…
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