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NEET PHYSICSEasy

Light with an energy flux of 25×104 W m−225 \times 10^4 \text{ W m}^{-2} falls on a perfectly reflecting surface at normal incidence. If the surface area is 15 cm215 \text{ cm}^2, the average force exerted on the surface is:

A

1.25×10−6 N1.25 \times 10^{-6} \text{ N}

B

2.50×10−6 N2.50 \times 10^{-6} \text{ N}

C

1.20×10−6 N1.20 \times 10^{-6} \text{ N}

D

3.0×10−6 N3.0 \times 10^{-6} \text{ N}

Step-by-Step Solution

  1. Identify the Formula: The force exerted by an electromagnetic wave on a surface depends on whether the surface absorbs or reflects the radiation. For a perfectly reflecting surface at normal incidence, the change in momentum is double that of an absorbing surface. The force (FF) is given by: F=2IAcF = \frac{2IA}{c} where II is the energy flux (intensity), AA is the surface area, and cc is the speed of light .

  2. Convert Units:

  • Energy Flux (II) = 25×104 W/m225 \times 10^4 \text{ W/m}^2
  • Surface Area (AA) = 15 cm2=15×10−4 m215 \text{ cm}^2 = 15 \times 10^{-4} \text{ m}^2
  • Speed of light (cc) ≈3×108 m/s\approx 3 \times 10^8 \text{ m/s}
  1. Calculate Force: Substitute the values into the formula: F=2×(25×104)×(15×10−4)3×108F = \frac{2 \times (25 \times 10^4) \times (15 \times 10^{-4})}{3 \times 10^8} F=2×25×15×1003×108F = \frac{2 \times 25 \times 15 \times 10^0}{3 \times 10^8} F=7503×108F = \frac{750}{3 \times 10^8} F=250×10−8 NF = 250 \times 10^{-8} \text{ N} F=2.50×10−6 NF = 2.50 \times 10^{-6} \text{ N}
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