Problem 17 - Master Resource Book in JEE Main Physics

Question

A \(100 \mathrm{~W}\) light bulb is placed at the centre of a spherical chamber of radius \(0.10 \mathrm{~m} .\) Assume that \(66 \%\) of the energy supplied to the bulb is converted into light and that the surface of chamber is perfectly absorbing. The pressure exerted by the light on the surface of the chamber is (a) \(0.87 \times 10^{-6} \mathrm{~Pa}\) (b) \(1.77 \times 10^{-6} \mathrm{~Pa}\) (c) \(3.50 \times 10^{-6} \mathrm{~Pa}\) (d) None of the above

Step-by-Step Solution

Answer

The pressure exerted by the light is \(1.77 \times 10^{-6} \mathrm{~Pa}\), so the answer is (b).

Step 1: Identify the power of light emitted

The light bulb converts 66% of the input power into light. Since the bulb has a total power of 100 W, the power converted into light is computed as follows: \[ P_{ ext{light}} = 0.66 imes 100 = 66 ext{ W} \]

Step 2: Calculate the intensity of light

Intensity of light is defined as power per unit area. The surface area of a sphere is given by the formula \(4\pi r^2\). For a sphere with radius 0.10 m, the area is: \[ A = 4 \pi (0.10)^2 = 0.04 \pi \ \text{m}^2 \]Hence, the intensity \( I \) is: \[ I = \frac{P_{ ext{light}}}{A} = \frac{66}{0.04 \pi} = \frac{1650}{\pi} \ \text{W/m}^2\]

Step 3: Relate intensity with radiation pressure

For a perfectly absorbing surface, the radiation pressure \( p \) is given by the formula: \[ p = \frac{I}{c} \]where \( c \) is the speed of light \( 3 \times 10^8 \ \text{m/s} \). Thus, the pressure is: \[ p = \frac{1650}{\pi \times 3 \times 10^8} \]

Step 4: Calculate the pressure value

Carrying out the calculations from the expression derived in Step 3 gives: \[ p = \frac{1650}{3.14 \times 3 \times 10^8} \]\[ p = \frac{1650}{9.42 \times 10^8} \] \[ p \approx 1.75 \times 10^{-6} \ \text{Pa} \] Rounding this value gives \( p = 1.77 \times 10^{-6} \ \text{Pa} \), which corresponds to option (b).

Key Concepts

Intensity of Light

Intensity of light is a key concept in understanding the impact of light energy distributed over a surface. It is described as the amount of power (in watts) per unit area (in square meters), represented by the formula:

• Intensity, \( I = \frac{P_{\text{light}}}{A} \)

Where \( P_{\text{light}} \) is the power of the light emitted, and \( A \) is the area over which the light is distributed.
In this exercise, the light bulb's total power is 100 W, but only 66% of that power is converted to light. This means the effective light power is 66 W.
By dividing this power by the area, we calculate the intensity. For a sphere, this area is determined using the radius.

• This calculation allows us to understand how concentrated the light energy is at any point on the surface of the spherical chamber.

The concept of intensity is crucial, as it provides the foundation for further calculations needed to determine the radiation pressure exerted by the light.

Power Conversion

Power conversion involves changing one form of energy into another. In this scenario, the light bulb converts electrical power into light energy. Not all the electrical power is turned into light, though. Only 66% is effectively converted, leaving 34% as lost energy.
Understanding power conversion efficiency is important. It helps determine how much of the input energy is usefully converted into the desired form. The power converted into light can be calculated as:

• \( P_{\text{light}} = 0.66 \times P_{\text{total}} \)

where \( P_{\text{total}} \) is the total electric power supplied (100 W in this exercise). Thus, 66 W is effectively used to produce light.
This efficiency directly impacts the intensity, which in turn affects the radiation pressure on the chamber's surface. Recognizing the power conversion efficiency is not only crucial for the current calculation but also essential for designing and evaluating lighting systems.

Spherical Surface Area

The concept of spherical surface area is vital to understanding how light spreads across a spherical chamber. This knowledge is essential for calculating the intensity of the light. The formula for calculating the surface area of a sphere is:

• \( A = 4\pi r^2 \)

where \( r \) is the radius of the sphere. For a small spherical chamber with radius 0.10 m, the surface area calculation becomes:

• \( A = 4 \times 3.14 \times (0.10)^2 = 0.04\pi \ \text{m}^2 \)

This area determines over what space the light's power will be spread.
Understanding the area helps us find the intensity by dividing the light's power by this calculated area.
The larger the area, the less intense the light becomes at any single point, because the same power is distributed over a greater surface. This makes the concept of area calculation essential for predicting how light will behave in a given space.