Question

The mass of mirror is 20g and the energy passed by Lens is only 30% of total incident on the lens as shown. Calculate the power of the source in order to balance the weight of mirror?

• A. $10^8w$
• B. 2 x $10^8w$
• C. $\frac{1}{2} \times 10^8w$
• D. 0.2 x $10^8w$

Answer 1

Answer: (A)

$10^8w$

Answer 2

Here's how to calculate the power of the source required to balance the weight of the mirror:

1. Calculate the weight of the mirror: $W = mg = 0.02 \, \text{kg} \times 10 \, \text{m/s}^2 = 0.2 \, \text{N}$

2. Determine the force required to balance the weight: The upward force due to radiation pressure must equal the weight of the mirror, so $F = W = 0.2 \, \text{N}$.

3. Relate the force due to radiation pressure to the power incident on the mirror: For a perfectly reflecting mirror, the force is given by $F = \frac{2P_{\text{incident on mirror}}}{c}$, where $c$ is the speed of light ($3 \times 10^8 \, \text{m/s}$).

4. Calculate the power incident on the mirror: $P_{\text{incident on mirror}} = \frac{Fc}{2} = \frac{0.2 \, \text{N} \times 3 \times 10^8 \, \text{m/s}}{2} = 0.3 \times 10^8 \, \text{W}$

5. Relate the power incident on the mirror to the power of the source: The problem states that the power incident on the mirror is 30% of the power incident on the lens. We assume the power of the source (P) is the power incident on the lens. Therefore, $P_{\text{incident on mirror}} = 0.30 \times P$.

6. Calculate the power of the source: $P = \frac{P_{\text{incident on mirror}}}{0.30} = \frac{0.3 \times 10^8 \, \text{W}}{0.30} = 10^8 \, \text{W}$

Therefore, the power of the source required to balance the weight of the mirror is $10^8 \, \text{W}$.