Question

A laser beam ( 2 = 633 nm) has an power of 3 mW. What will be the pressure exerted on a surface by this beam if the cross sectional area is 3 mm(Assume perfect reflection and normal incidence) (1) 6.6x10-3 N/m2 (2) 6.6x10-6 N/m2 (3) 6.6x10-9 N/m2 (4) 6.6 N/m2​

Answer 1

Answer:

the pressure exerted on a surface by this beam is

$P = 6.6 \times 10^{-6} Pa$

Explanation:

As we know that the surface is 100% reflecting

So here the radiation pressure is given by the formula

$P = \frac{2I}{c}$

here we know that

$I = \frac{Power}{Area}$

so we have

$P = 3 mW$

$A = 3\times 10^{-6} m^2$

now we have

$I = \frac{3\times 10^{-3}}{3 \times 10^{-6}}$

$I = 1000 W/m^2$

now the radiation pressure is given as

$P = \frac{2\times 1000}{3 \times 10^8}$

$P = 6.6 \times 10^{-6} Pa$

#Learn

Topic : Radiation pressure

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