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NEET
Exam
NEET
Subject
Physics
Chapter
Electromagnetic Waves

NEET · Physics · Electromagnetic Waves

A point source of light having a power of 1000π W radiates...

A point source of light having a power of 1000π W1000\pi \text{ W} radiates isotropically in all directions. A small detector having an area of 10 cm210 \text{ cm}^2 is placed at a distance of 5 m5 \text{ m} from the source, such that the light falls normally on it. The total energy received by the detector in 1 minute1 \text{ minute} is:

Options

  1. A6000 J6000 \text{ J}
  2. B0.01 J0.01 \text{ J}
  3. C2.4 J2.4 \text{ J}
  4. D0.6 J0.6 \text{ J}

Correct answer

D. 0.6 J0.6 \text{ J}

Step-by-step solution

The intensity of light at a distance rr from a point source is given by:

I=P4πr2I = \dfrac{P}{4\pi r^2}

Substituting the given values:

I=1000π4π(5)2=1000100=10 W m−2I = \dfrac{1000\pi}{4\pi (5)^2} = \dfrac{1000}{100} = 10 \text{ W m}^{-2}

The energy received by the detector is:

E=I×A×tE = I \times A \times t

Converting the given area and time into SI units:
A=10 cm2=10×10−4 m2=10−3 m2A = 10 \text{ cm}^2 = 10 \times 10^{-4} \text{ m}^2 = 10^{-3} \text{ m}^2
t=1 minute=60 st = 1 \text{ minute} = 60 \text{ s}

Calculating the energy:
E=10×10−3×60=0.6 JE = 10 \times 10^{-3} \times 60 = 0.6 \text{ J}

Answer: 0.6 J0.6 \text{ J}
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