JEE Mains · Physics · STD 12 - 10. Wave optics
The maximum intensity in a Young's double slit experiment is \(I_0\). Distance between the slits (\(d\)) is \(5\lambda\), where \(\lambda\) is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at \(D = 10d\) is _______.
- A \(\dfrac{I_0}{4}\)
- B \(\dfrac{I_0}{2}\)
- C \(I_0\)
- D \(\dfrac{3I_0}{4}\)
Answer & Solution
Correct Answer
(B) \(\dfrac{I_0}{2}\)
Step-by-step Solution
Detailed explanation
The position of the point on the screen exactly opposite to one of the slits is at a distance \(y = \dfrac{d}{2}\) from the central maximum. The path difference \(\Delta x\) at this point is given by: \(\Delta x = \dfrac{yd}{D}\) Substituting \(y = \dfrac{d}{2}\) and…
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