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Question: A plane electromagnetic wave of wave intensity ![](https://cdn.pureessence.tech/canvas_204.png?top_l...

A plane electromagnetic wave of wave intensity strikes a small mirror area 40 cm240 \mathrm {~cm} ^ { 2 } , held perpendicular to the approaching wave. The momentum transferred by the wave to the mirror each second will be

A

6.4×10−7 kg−m/s26.4 \times 10 ^ { - 7 } \mathrm {~kg} - \mathrm { m } / \mathrm { s } ^ { 2 }

B

4.8×10−8 kg−m/s24.8 \times 10 ^ { - 8 } \mathrm {~kg} - \mathrm { m } / \mathrm { s } ^ { 2 }

C

3.2×10−9 kg−m/s23.2 \times 10 ^ { - 9 } \mathrm {~kg} - \mathrm { m } / \mathrm { s } ^ { 2 }

D

1.6×10−10 kg−m/s21.6 \times 10 ^ { - 10 } \mathrm {~kg} - \mathrm { m } / \mathrm { s } ^ { 2 }

Answer

1.6×10−10 kg−m/s21.6 \times 10 ^ { - 10 } \mathrm {~kg} - \mathrm { m } / \mathrm { s } ^ { 2 }

Explanation

Solution

In one second p=2Uc=2SavAc=2×6×40×10−43×108p = \frac { 2 U } { c } = \frac { 2 S _ { a v } A } { c } = \frac { 2 \times 6 \times 40 \times 10 ^ { - 4 } } { 3 \times 10 ^ { 8 } }

=1.6×10−10 kg−m/s2= 1.6 \times 10 ^ { - 10 } \mathrm {~kg} - \mathrm { m } / \mathrm { s } ^ { 2 }