MHT CET · Physics · Wave Optics
In Young's double slit experiment, the intensity on screen at a point, where path difference is \(\frac{\lambda}{4}\) is \(\frac{K}{4}\). The intensity at a point when path difference is ' \(\lambda\) ' will be \(\left[\cos \frac{\pi}{2}=0, \cos 2 \pi=1\right]\)
- A 4 K
- B 2 K
- C K
- D \(\frac{K}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{K}{2}\)
Step-by-step Solution
Detailed explanation
\(\phi_1 = \frac{2\pi}{\lambda} \left(\frac{\lambda}{4}\right) = \frac{\pi}{2}\) \(I_1 = I_{max} \cos^2\left(\frac{\pi/2}{2}\right) = I_{max} \cos^2\left(\frac{\pi}{4}\right) = I_{max} \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_{max}}{2}\)
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