Question

A parallel beam of light is incident normally on a plane surface absorbing \(40\%\) of light and reflecting the rest. If the incident beam carries \(60\) W of power the force exerted by it on the surface is (C = velocity of light = \(3\times 10^{8}\) m/s )

• A. \(3.2\times 10^{-8}\) N
• B. \(3.2\times 10^{-7}\) N
• C. \(5.12\times 10^{-7}\) N
• D. \(5.12\times 10^{-8}\) N

Hint:

Force from absorbed light is P/c and from reflected light is 2P/c for the respective fractions.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Light carries momentum. A beam that is fully absorbed exerts force \(\dfrac{P}{c}\), and a beam that is fully reflected back exerts \(\dfrac{2P}{c}\).

Step 2: Split the Beam
Absorbed fraction \(= 0.4\), reflected fraction \(= 0.6\).
\(F = 0.4\dfrac{P}{c} + 0.6\dfrac{2P}{c} = 1.6\dfrac{P}{c}\)

Step 3: Calculation
\(\dfrac{P}{c} = \dfrac{60}{3\times 10^{8}} = 2\times 10^{-7}\) N
\(F = 1.6 \times 2\times 10^{-7} = 3.2\times 10^{-7}\) N

Step 4: Compare the options
The force is \(3.2\times 10^{-7}\) N, option (B).

Final Answer:
The force exerted on the surface is 3.2 x 10^-7 N. \[ \boxed{\text{(B) }3.2\times 10^{-7}\ \text{N}} \]