NEET · Physics · STD 12 - 10. Wave optics
In Young's double slit experiment, using monochromatic light of wavelength \(\lambda\), the intensity of light at a point on the screen where the path difference is \(\lambda\), is K units. The intensity of light at a point where the path difference is \(\frac{\lambda}{3}\) will be :
- A \(\frac{K}{4}\)
- B K
- C 2K
- D \(\frac{K}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{K}{4}\)
Step-by-step Solution
Detailed explanation
(A) \(\frac{K}{4}\)
Phase difference \(\phi\) is related to path difference \(\Delta x\) by \(\phi=\frac{2 \pi}{\lambda} \Delta x\).
The intensity \(I\) at a point on the screen is given by \(I=I_{\max } \cos ^2\left(\frac{\phi}{2}\right)\).
For \(\Delta x=\lambda, \phi=\frac{2 \pi}{\lambda} \times \lambda=2 \pi\).
Intensity \(I=I_{\text {max }} \cos ^2(\pi)=I_{\text {max }}=K\).
For \(\Delta x=\frac{\lambda}{3}, \phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}=\frac{2 \pi}{3}\).
Intensity \(I^{\prime}=I_{\max } \cos ^2\left(\frac{\pi}{3}\right)=K\left(\frac{1}{2}\right)^2=\frac{K}{4}\).
Phase difference \(\phi\) is related to path difference \(\Delta x\) by \(\phi=\frac{2 \pi}{\lambda} \Delta x\).
The intensity \(I\) at a point on the screen is given by \(I=I_{\max } \cos ^2\left(\frac{\phi}{2}\right)\).
For \(\Delta x=\lambda, \phi=\frac{2 \pi}{\lambda} \times \lambda=2 \pi\).
Intensity \(I=I_{\text {max }} \cos ^2(\pi)=I_{\text {max }}=K\).
For \(\Delta x=\frac{\lambda}{3}, \phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}=\frac{2 \pi}{3}\).
Intensity \(I^{\prime}=I_{\max } \cos ^2\left(\frac{\pi}{3}\right)=K\left(\frac{1}{2}\right)^2=\frac{K}{4}\).
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