COMEDK · Physics · 26. Wave Optics
In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is \(\lambda\) is \(\mathrm{K}\) units (\(\lambda\) is the wavelength of light used). The percentage change in intensity at a point where the path difference is \(\dfrac{\lambda}{6}\) and the above point is
- A 50%
- B 25%
- C 75%
- D 4%
Answer & Solution
Correct Answer
(B) 25%
Step-by-step Solution
Detailed explanation
The intensity \(I\) at any point on the screen in Young's double slit experiment is given by \(I = I_{max} \cos^{2}\left(\dfrac{\phi}{2}\right)\), where \(\phi\) is the phase difference. The phase difference \(\phi\) is related to the path difference \(\Delta x\) by…
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