Q.40. A point light source emits E.M. waves in free space. A detector, placed at a distance of L m,

Q: What is the correct answer for JEE Main Physics Q40?

The correct answer is B. The intensity of electromagnetic waves from a point source in free space follows an inverse square law with distance. The intensity I at a distance r from a point source is given by $I = \frac{P}{4\pi r^2}$, where P is the power of the source. Initially, the detector is at a distance L and measures intensity $I_0$. So, $I_0 = \frac{P}{4\pi L^2}$. The detector is then shifted to another location

Q: Q.40. A point light source emits E.M. waves in free space. A detector, placed at a distance of L m, measures the intensity as . The detector is now shifted to another location on the same spherical…

The correct answer is B. The intensity of electromagnetic waves from a point source in free space follows an inverse square law with distance. The intensity I at a distance r from a point source is given by $I = \frac{P}{4\pi r^2}$, where P is the power of the source. Initially, the detector is at a distance L and measures intensity $I_0$. So, $I_0 = \frac{P}{4\pi L^2}$. The detector is then shifted to another location

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

A. $I_0/4$ · B. $I_0$ · C. $I_0/2$ · D. $I_0/\sqrt{2}$

Question

Q.40. A point light source emits E.M. waves in free space. A detector, placed at a distance of L m, measures the intensity as . The detector is now shifted to another location on the same spherical surface ensuring the angle between original location and new location as 45°. The measured intensity at new location will be ____.

FirstInTest · Accepted Answer

The intensity of electromagnetic waves from a point source in free space follows an inverse square law with distance. The intensity I at a distance r from a point source is given by $I = \frac{P}{4\pi r^2}$, where P is the power of the source.

Initially, the detector is at a distance L and measures intensity $I_0$. So, $I_0 = \frac{P}{4\pi L^2}$.

The detector is then shifted to another location on the *same spherical surface*. This means the distance from the point source to the detector remains the same, i.e., $r = L$.

Since the distance L remains constant, and the power P of the source is constant, the intensity measured by the detector will also remain the same.
The angle between the original and new location (45°) is irrelevant because the intensity only depends on the distance from the point source in free space, not the angular position on a sphere of constant radius.

Therefore, the measured intensity at the new location will still be $I_0$.

Answer

$I_0/4$

Answer

$I_0$

Answer

$I_0/2$

Answer

$I_0/\sqrt{2}$

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