MHT CET · Physics · Wave Optics
In Young's double slit experiment, the resultant intensity of light at a point on the screen is 'I' when the path difference is \(\lambda^{\prime}\) '. When the path difference is \(\frac{\lambda}{4}\), the intensity at a point will be \(\left(\lambda=\right.\) wavelength of light, \(\cos 180^{\circ}=-1, \cos 45^{\circ}=\frac{1}{\sqrt{2}}\) )
- A Zero
- B I
- C \(\frac{\mathrm{I}}{2}\)
- D \(\frac{\mathrm{I}}{4}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{I}}{2}\)
Step-by-step Solution
Detailed explanation
\(\phi_1 = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi\) \(I = 4I_0 \cos^2\left(\frac{\phi_1}{2}\right) = 4I_0 \cos^2(\pi) = 4I_0\)
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