Q.42. The maximum intensity in a Young's double slit experiment is . Distance between the slits (d)

Q: What is the correct answer for JEE Main Physics Q42?

The correct answer is D. In Young's Double Slit Experiment, the intensity at a point is given by $I = I_0 \cos^2(\frac{\phi}{2})$, where $\phi$ is the phase difference. The phase difference is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$. The screen is at a distance D = 10d. The distance between slits is d = $5\lambda$. We need to find the intensity exactly opposite to one of the sli

Q: Q.42. The maximum intensity in a Young's double slit experiment is . Distance between the slits (d) is , where is the wavelength of light used. The intensity of the fringe, exactly opposite to one…

The correct answer is D. In Young's Double Slit Experiment, the intensity at a point is given by $I = I_0 \cos^2(\frac{\phi}{2})$, where $\phi$ is the phase difference. The phase difference is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$. The screen is at a distance D = 10d. The distance between slits is d = $5\lambda$. We need to find the intensity exactly opposite to one of the sli

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

A. $I_0/4$ · B. $I_0/2$ · C. $3I_0/4$ · D. $I_0/2$

Question

Q.42. The maximum intensity in a Young's double slit experiment is . Distance between the slits (d) is , where  is the wavelength of light used. The intensity of the fringe, exactly opposite to one of the slits on the screen, placed at D = 10 d is

FirstInTest · Accepted Answer

In Young's Double Slit Experiment, the intensity at a point is given by $I = I_0 \cos^2(\frac{\phi}{2})$, where $\phi$ is the phase difference. The phase difference is related to the path difference $\Delta x$ by $\phi = \frac{2\pi}{\lambda} \Delta x$.
The screen is at a distance D = 10d. The distance between slits is d = $5\lambda$. We need to find the intensity exactly opposite to one of the slits. Let's assume the slits are at $y = -d/2$ and $y = +d/2$. The point on the screen is at $y = d/2$ (opposite one slit).
The path difference $\Delta x = \frac{yd}{D}$. In this case, $y = d/2$. So, $\Delta x = \frac{(d/2)d}{D} = \frac{d^2}{2D}$.
Substitute $d = 5\lambda$ and $D = 10d = 10(5\lambda) = 50\lambda$.
$\Delta x = \frac{(5\lambda)^2}{2(50\lambda)} = \frac{25\lambda^2}{100\lambda} = \frac{\lambda}{4}$.
Now, calculate the phase difference: $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \left( \frac{\lambda}{4} ight) = \frac{\pi}{2}$.
Finally, calculate the intensity: $I = I_0 \cos^2(\frac{\phi}{2}) = I_0 \cos^2(\frac{\pi/2}{2}) = I_0 \cos^2(\frac{\pi}{4})$.
Since $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$, then $\cos^2(\frac{\pi}{4}) = \left( \frac{1}{\sqrt{2}} ight)^2 = \frac{1}{2}$.
So, $I = I_0 	imes \frac{1}{2} = \frac{I_0}{2}$.

The solution provided uses a different approach for path difference: $\Delta x = \frac{d}{2} \frac{d}{D}$. This is incorrect. The path difference for a point y from the center is $\frac{yd}{D}$. If the point is opposite one slit, its distance from the center is $d/2$. So $\Delta x = \frac{(d/2)d}{D}$. The solution's calculation of $\Delta x = \frac{d^2}{2D}$ is correct. Then it calculates $\Delta x = \frac{25\lambda^2}{2(10d)} = \frac{25\lambda^2}{2(10 	imes 5\lambda)} = \frac{25\lambda^2}{100\lambda} = \frac{\lambda}{4}$. This is also correct. Then $\phi = \frac{2\pi}{\lambda} \frac{\lambda}{4} = \frac{\pi}{2}$. Then $I = I_0 \cos^2(\frac{\pi/2}{2}) = I_0 \cos^2(\frac{\pi}{4}) = I_0 (\frac{1}{\sqrt{2}})^2 = \frac{I_0}{2}$. The solution's final answer is $I_0/2$, which corresponds to option (4) in the provided image (which is $I_0/2$). The OCR might have misread $I_0/2$ as $3I_0/4$. Let's assume the option is $I_0/2$. If the option is $3I_0/4$, then there is a mismatch. Given the calculation, $I_0/2$ is the correct answer.

Answer

$I_0/4$

Answer

$I_0/2$

Answer

$3I_0/4$

Answer

$I_0/2$

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