KCET · Physics · Dual Nature of Matter
A light beam of intensity \(20 \mathrm{~W} / \mathrm{cm}^{2}\) is incident normally on a perfectly reflecting surface of sides \(25 \mathrm{~cm} \times 15 \mathrm{~cm}\). The momentum imparted to the surface by the light per second is
- A \(2 \times 10^{-5} \mathrm{~kg}-\mathrm{ms}^{-1}\)
- B \(1 \times 10^{-5} \mathrm{~kg}-\mathrm{ms}^{-1}\)
- C \(5 \times 10^{-5} \mathrm{~kg}-\mathrm{ms}^{-1}\)
- D \(1.2 \times 10^{-5} \mathrm{~kg}-\mathrm{ms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(5 \times 10^{-5} \mathrm{~kg}-\mathrm{ms}^{-1}\)
Step-by-step Solution
Detailed explanation
Given, intensity, \(I=20 \mathrm{~W} / \mathrm{cm}^{2}\)
Area of surface, \(A=(25 \times 15) \mathrm{cm}^{2}\)
and time, \(t=1 \mathrm{~s}\)
The intensity of light of energy \(E\) is given by
\(\begin{aligned} I &=\frac{E}{A t} \Rightarrow 20=\frac{E}{25 \times 15 \times 1} \\ \Rightarrow \quad E &=20 \times 25 \times 15 \mathrm{~J} \end{aligned}\)
The momentum imparted to a perfectly reflecting surface is given by
\(p=\frac{2 E}{c}=\frac{2 \times 20 \times 25 \times 15}{3 \times 10^{8}}=5 \times 10^{-5} \mathrm{~kg} \mathrm{~ms}^{-1}\)
Area of surface, \(A=(25 \times 15) \mathrm{cm}^{2}\)
and time, \(t=1 \mathrm{~s}\)
The intensity of light of energy \(E\) is given by
\(\begin{aligned} I &=\frac{E}{A t} \Rightarrow 20=\frac{E}{25 \times 15 \times 1} \\ \Rightarrow \quad E &=20 \times 25 \times 15 \mathrm{~J} \end{aligned}\)
The momentum imparted to a perfectly reflecting surface is given by
\(p=\frac{2 E}{c}=\frac{2 \times 20 \times 25 \times 15}{3 \times 10^{8}}=5 \times 10^{-5} \mathrm{~kg} \mathrm{~ms}^{-1}\)
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