JEE Mains · Physics · STD 12 - 10. Wave optics
In a Young's double slit experiment, the intensity at some point on the screen is found to be \(\dfrac{3}{4}\) times of the maximum of the interference pattern. The path difference between the interfering waves at this point is \(\dfrac{\lambda}{x}\) where \(\lambda\) is wavelength of the incident light. The value of \(x\) is _______.
- A 1
- B 3
- C 5
- D 6
Answer & Solution
Correct Answer
(D) 6
Step-by-step Solution
Detailed explanation
The intensity at a point in Young's double slit experiment is given by \(I = I_{max} \cos^2\left(\dfrac{\phi}{2}\right)\) Given \(I = \dfrac{3}{4} I_{max}\) \(\dfrac{3}{4} I_{max} = I_{max} \cos^2\left(\dfrac{\phi}{2}\right)\)…
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