Q.36. In interference experiment the path difference between two interfering waves at a point A on t

Q: What is the correct answer for JEE Main Physics Q36?

The correct answer is C. Sol. The intensity $I$ at a point in an interference pattern is given by $I = I_0 \cos^2(\phi/2)$, where $I_0$ is the maximum intensity and $\phi$ is the phase difference. The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$. At point A: Path difference $\Delta x_A = \lambda/3$. Phase difference $\phi_A = \frac{2\pi}{\lambda} (\lambda/3

Q: Q.36. In interference experiment the path difference between two interfering waves at a point A on the screen is , where is the wavelength of these waves, and at another point B the path difference…

The correct answer is C. Sol. The intensity $I$ at a point in an interference pattern is given by $I = I_0 \cos^2(\phi/2)$, where $I_0$ is the maximum intensity and $\phi$ is the phase difference. The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$. At point A: Path difference $\Delta x_A = \lambda/3$. Phase difference $\phi_A = \frac{2\pi}{\lambda} (\lambda/3

Q: What are the options for this JEE Main Physics question?

A. 3 · B. 4 · C. 1/3 · D. 1/4

Question

Q.36. In interference experiment the path difference between two interfering waves at a point A on the screen is , where  is the wavelength of these waves, and at another point B the path difference is . The ratio of intensities at points A and B is ________.

FirstInTest · Accepted Answer

Sol. The intensity $I$ at a point in an interference pattern is given by $I = I_0 \cos^2(\phi/2)$, where $I_0$ is the maximum intensity and $\phi$ is the phase difference. The phase difference $\phi$ is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$. At point A: Path difference $\Delta x_A = \lambda/3$. Phase difference $\phi_A = \frac{2\pi}{\lambda} (\lambda/3) = \frac{2\pi}{3}$. Intensity $I_A = I_0 \cos^2(\phi_A/2) = I_0 \cos^2(\frac{2\pi}{3 \cdot 2}) = I_0 \cos^2(\pi/3) = I_0 (1/2)^2 = \frac{I_0}{4}$. At point B: Path difference $\Delta x_B = \lambda/6$. Phase difference $\phi_B = \frac{2\pi}{\lambda} (\lambda/6) = \frac{\pi}{3}$. Intensity $I_B = I_0 \cos^2(\phi_B/2) = I_0 \cos^2(\frac{\pi}{3 \cdot 2}) = I_0 \cos^2(\pi/6) = I_0 (\frac{\sqrt{3}}{2})^2 = I_0 (3/4) = \frac{3I_0}{4}$. The ratio of intensities at points A and B is $\frac{I_A}{I_B} = \frac{I_0/4}{3I_0/4} = \frac{1}{3}$.

Answer

3

Answer

4

Answer

1/3

Answer

1/4

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