A point light source S emits electromagnetic waves in free space.

A detector placed at point A, which is 1 m from the source, measures an intensity I₀.

The detector is then moved to point B as shown in the figure, where angle ASB = 45°.

Find the intensity measured at point B.

Options:

• (1) I₀
• (2) I₀/4
• (3) I₀/√2
• (4) I₀/2

Intensity due to a point source follows the inverse square law:

I ∝ 1/r²

Distance SA = 1 m

Intensity at A:

I_A = I₀

From the figure:

• Triangle SAB is right-angled at A.
• Angle S = 45°.

Therefore the triangle is an isosceles right triangle.

Hence:

SA = AB = 1 m

Distance:

SB = √(1² + 1²)

= √2 m

Applying inverse square law:

I_B = I₀ (1²)/(√2)²

= I₀/2

Final Answer

Option (4) I₀/2