Question

The intensity at spherical surface due to an isotropic point source placed at its center is $I_0$. If its volume is increased by $8$ times, what will be intensity at the spherical surface? 

• A. Increase by $128$ times
• B. Increase by $8$ times
• C. Decrease by $4$ times
• D. Decrease by $8$ times

Hint:

For isotropic sources, intensity varies inversely as square of distance.

Answer 1

The Correct Option is C

Step 1: Relation between intensity and surface area

For an isotropic point source, the intensity of light at the surface of a sphere is given by:

I = P / A

where
P = power of the source (constant)
A = surface area of the sphere

Surface area of a sphere:

A = 4πr²

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Step 2: Use the given change in volume

Volume of a sphere is:

V = (4/3)πr³

Given that the volume increases by 8 times:

V₁ / V₀ = 8

⇒ (r₁/r₀)³ = 8

⇒ r₁ = 2r₀

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Step 3: Compare the surface areas

Initial surface area:

A₀ = 4πr₀²

Final surface area:

A₁ = 4π(2r₀)² = 16πr₀²

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Step 4: Compare the intensities

Since intensity is inversely proportional to surface area:

I₁ / I₀ = A₀ / A₁

I₁ / I₀ = (4πr₀²) / (16πr₀²) = 1/4

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Final Answer:

The intensity of light decreases by
4 times