Question

14.
(3) II, 111
(4) 1, iv a
A source of power 1.2 MW emits photon which 14
incident on a surface 2.5 m away out of which
50% is reflected back what is the radiation
pressure on the surface ?
(1) 3.9 ~ 10 Pa/ / (2) 7.8 x 10- Pa
(3) 11.7 x 10Pa (4) 15.6 10 Pa
5.
A vessel with water is placed on a weighing pan
1110 and​

Answer 1

Radiation pressure = 7.6 × ${10^-^5}$ Pa.

Given:

A source of power = p = 1.2 MW = 1200000 W

Incident on a surface = r = 2.5 m

50% is reflected back = R =0.5

Find:

Pressure on the surface.

Formula used:

Radiation pressure = $\frac{I}{C}$ (1+r)

Intensity = I = $\frac{P}{4\pi r^{2} }$

  Where C = velocity of light = 3×$10^{8}$ m/s

Explanation:

Intensity = I = $\frac{P}{4\pi r^{2} }$

                 I =  $\frac{1200000}{4\pi 2.5^{2} }$

                 I = 1.52 ×$10^{4}$ W/m²

Radiation pressure = $\frac{I}{C}$ (1+R)

                                = $\frac{1.52 \times 10^4}{3\times 10^{8} }$ × (1+0.5)

Radiation pressure = 7.6 × $10^-^5$ Pa.

To learn more....

1)If wave number of radiation is 20 m-1 then frequency of the radiation will be.

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2)Radiation power of 3.31 watt is falling on a surface .if 20% of light is absorbed and remaining is reflected back,calculate number of photons reflected per second.take the frequency of each photon 10*12 hz.

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