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NEET PHYSICSELECTROMAGNETIC WAVESEasy

Question

Light with an energy flux of 25×104 W m225 \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

1.25×106 N1.25 \times 10^{-6} \text{ N}

B

2.50×106 N2.50 \times 10^{-6} \text{ N}

C

1.20×106 N1.20 \times 10^{-6} \text{ N}

D

3.0×106 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=15×104 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)×(15×104)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=250×108 NF = 250 \times 10^{-8} \text{ N} F=2.50×106 NF = 2.50 \times 10^{-6} \text{ N}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROMAGNETIC WAVES. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSELECTROMAGNETIC WAVESenergyperfectlyreflectingsurfacenormal

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