Question

A point source is kept at the center of a spherically enclosed detector. If the volume of the detector is increased by 8 times, the intensity will

• A. increase by 8 times
• B. increase by 64 times
• C. decrease by 4 times
• D. decrease by 8 times

Hint:

For point sources, intensity always follows the inverse square law.

Answer 1

The Correct Option is C

To solve this problem, we need to understand how the intensity of light from a point source changes with the enclosure's dimensions.

Concept Explanation:

The intensity (I) of light from a point source is defined as the power per unit area. Mathematically, it is given by:

\(I = \frac{P}{A}\)

where P is the power of the source and A is the area over which the power is spread.

For a spherical detector, the surface area (A) is calculated as:

\(A = 4\pi r^2\), where \(r\) is the radius of the sphere.

Given that the volume of the sphere is increased by 8 times, we are required to find the new intensity.

Volume and Radius Relationship:

The volume (V) of a sphere is given by:

\(V = \frac{4}{3}\pi r^3\)

If the volume increases by a factor of 8, then:

\(8V = \frac{4}{3}\pi (r_{\text{new}})^3\)

where \(r_{\text{new}}\) is the new radius.

From \(V = \frac{4}{3}\pi r^3\), we get:

\(r_{\text{new}} = \sqrt[3]{8} \cdot r = 2r\)

Calculating the New Intensity:

Since the radius is doubled, the new surface area becomes:

\(A_{\text{new}} = 4\pi (2r)^2 = 4\pi \cdot 4r^2 = 16\pi r^2 = 4A\)

The intensity depends inversely on the area. Therefore, when the area is quadrupled, the intensity is reduced by a factor of 4.

Thus, the new intensity is \(I_{\text{new}} = \frac{I}{4}\).

Conclusion:

Therefore, when the volume of the detector is increased by 8 times, the intensity decreases by 4 times.

Correct Answer: decrease by 4 times