Question:
In YDSE, intensity of both coherent sources are \(I_0\). Separation between the two slits is 5 cm and screen is placed at a distance of 50 cm from the slit. Wavelength of light is 6000 \(\text{Å}\). If intensity for a point on the screen is \(I_0\), then find the path difference at point P.
In YDSE, intensity of both coherent sources are \(I_0\). Separation between the two slits is 5 cm and screen is placed at a distance of 50 cm from the slit. Wavelength of light is 6000 \(\text{Å}\). If intensity for a point on the screen is \(I_0\), then find the path difference at point P.
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In YDSE, for constructive interference, the path difference is an integer multiple of the wavelength. For maximum intensity, the path difference is \(\frac{\lambda}{2}\).
Updated On: Apr 7, 2026
- 1000 \(\text{Å}\)
- 2000 \(\text{Å}\)
- 3000 \(\text{Å}\)
- 5000 \(\text{Å}\)
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The Correct Option is B
Solution and Explanation
In Young’s Double Slit Experiment (YDSE), the intensity at a point on the screen is given by:
\[
I = I_0 \cos^2\left(\frac{\pi \Delta x}{\lambda}\right)
\]
where:
- \(I_0\) is the maximum intensity,
- \(\Delta x\) is the path difference,
- \(\lambda\) is the wavelength of light.
We are given that the intensity is \(I_0\), so: \[ I_0 = I_0 \cos^2\left(\frac{\pi \Delta x}{\lambda}\right) \] This implies: \[ \cos^2\left(\frac{\pi \Delta x}{\lambda}\right) = 1 \] Thus: \[ \frac{\pi \Delta x}{\lambda} = \frac{1}{2} \quad \text{(for maximum intensity at the point P)} \] Solving for \(\Delta x\): \[ \Delta x = \frac{\lambda}{2\pi} \times \pi = \frac{\lambda}{2} \] Substitute the value of \(\lambda = 6000 \, \text{Å}\): \[ \Delta x = \frac{6000}{2} = 3000 \, \text{Å} \] Thus, the path difference at point P is \(3000 \, \text{Å}\). Final Answer: Option (C) 3000 \(\text{Å}\).
- \(I_0\) is the maximum intensity,
- \(\Delta x\) is the path difference,
- \(\lambda\) is the wavelength of light.
We are given that the intensity is \(I_0\), so: \[ I_0 = I_0 \cos^2\left(\frac{\pi \Delta x}{\lambda}\right) \] This implies: \[ \cos^2\left(\frac{\pi \Delta x}{\lambda}\right) = 1 \] Thus: \[ \frac{\pi \Delta x}{\lambda} = \frac{1}{2} \quad \text{(for maximum intensity at the point P)} \] Solving for \(\Delta x\): \[ \Delta x = \frac{\lambda}{2\pi} \times \pi = \frac{\lambda}{2} \] Substitute the value of \(\lambda = 6000 \, \text{Å}\): \[ \Delta x = \frac{6000}{2} = 3000 \, \text{Å} \] Thus, the path difference at point P is \(3000 \, \text{Å}\). Final Answer: Option (C) 3000 \(\text{Å}\).
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