Question

A 100 W light bulb is placed at the centre of a spherical chamber of radius 20 cm. Assume that 60% of the energy supplied to the bulb is converted into light and that the surface of the chamber is perfectly absorbing. Find the pressure exerted by the light on the surface of the chamber.

Answer 1

The pressure exerted by the light on the surface of the chamber is $4 \times 10^{-7} \mathrm{N} / \mathrm{m} 2$

Explanation:

Given data in the question  

Power of the Bulb light, P = 100 W

Spherical Chamber Length, R = 20 cm = 0.2 m

Since it transforms 60 per cent of the energy supply to the bulb into electricity.

The intensity of light reflected by the light is therefore, P' = 60 W

Step 1:

Force is  

$F=\frac{P}{c}$

Where c is Light speed

$F=\frac{60}{\left(3 \times 10^{8}\right)}=2 \times 10^{-7} N$

Step 2:

  $Pressure =\frac{ force} {area}$

$\begin{aligned}=\frac{2 \times 10^{-7}}{\left(4 \times 3.14 \times(0.2)^{2}\right)}(A=4 \pi r 2) \\=& \frac{1}{(8 \times 3.14)} \times 10^{-5} \\=& 0.039 \times 10^{-5} \\&=3.9 \times 10^{-7} \\=& 4 \times 10^{-7} N / \mathrm{m} 2\end{aligned}$