A light beam of intensity 20 W/cm2 is incident normally on a perfectly reflecting surface of sides 25 cm × 15 cm. The momentum imparted to the surface by the light per second is

Question

A light beam of intensity 20 W/cm2 is incident normally on a perfectly reflecting surface of sides 25 cm × 15 cm. The momentum imparted to the surface by the light per second is (A) 2×10-5 kg ms-1 (B) 1×10-5 kg ms-1 (C) 5×10-5 kg ms-1 (D) 1.2×10-5 kg ms-1

Tardigrade Admin · Accepted Answer

Intensity I = E / A where E is energy of radiation and A is incident area ⇒ E = IA Momentum of radiation is given by P =2 E / c =2 IA / c Where c is speed of light. P=(2 × 20 × 104 ( W / m 2) × 375 × 10-4 m 2/3 × 108 m / s ) =5 × 10-5 kg ms -1

Q. A light beam of intensity $20 W / c m^{2}$ is incident normally on a perfectly reflecting surface of sides $25 c m \times 15 c m$ . The momentum imparted to the surface by the light per second is

$2 \times 1 0^{- 5} k g m s^{- 1}$

$1 \times 1 0^{- 5} k g m s^{- 1}$

$5 \times 1 0^{- 5} k g m s^{- 1}$

$1.2 \times 1 0^{- 5} k g m s^{- 1}$

Q. A light beam of intensity 20W/cm2 is incident normally on a perfectly reflecting surface of sides 25cm×15cm. The momentum imparted to the surface by the light per second is

2×10−5kgms−1

1×10−5kgms−1

5×10−5kgms−1

1.2×10−5kgms−1

Intensity I=E/A where E is energy of radiation and A is incident area ⇒E=IA Momentum of radiation is given by P=2E/c=2IA/c Where c is speed of light. P=3×108m/s2×20×104m2W​×375×10−4m2​ =5×10−5kgms−1

Q. A light beam of intensity $20 W / c m^{2}$ is incident normally on a perfectly reflecting surface of sides $25 c m \times 15 c m$ . The momentum imparted to the surface by the light per second is

$2 \times 1 0^{- 5} k g m s^{- 1}$

$1 \times 1 0^{- 5} k g m s^{- 1}$

$5 \times 1 0^{- 5} k g m s^{- 1}$

$1.2 \times 1 0^{- 5} k g m s^{- 1}$