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2 years ago

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A black, totally absorbing piece of cardboard of area A = 1.7 cm2 intercepts light with an intensity of 8.1 W/m2 from a camera s

trobe light. What radiation pressure is produced on the cardboard by the light?

2 answers:

Ipatiy [6.2K]2 years ago

6 0

Answer:

The answer is 2.7x10⁻⁸ Pa

Explanation:

The expression for total absorption is:

Pr = I/c

Where I = intensity of light

c = velocity of light

Replacing:

Pr = 8.1/3x10⁸ = 2.7x10⁻⁸ Pa

Furkat [3]2 years ago

3 0

Answer:

2.7x10⁻⁸ N/m²

Explanation:

Since the piece of cardboard absorbs totally the light, the radiation pressure can be found using the following equation:

$p_{rad} = \frac{I}{c}$

<u>Where:</u>

$p_{rad}$ : is the radiation pressure

I: is the intensity of the light = 8.1 W/m²

c: is the speed of light = 3.00x10⁸ m/s

Hence, the radiation pressure is:

$p_{rad} = \frac{I}{c} = \frac{8.1 W/m^{2}}{3.00 \cdot 10^{8} m/s} = 2.7 \cdot 10^{-8} N/m^{2}$

Therefore, the radiation pressure that is produced on the cardboard by the light is 2.7x10⁻⁸ N/m².

I hope it helps you!

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