Light of uniform intensity impinges perpendicularly on a totally refle

Light of uniform intensity impinges perpendicularly on a totally reflecting surface. If the area of the surface is halved, the radiation force on it will become

[amp_mcq option1=”double” option2=”half” option3=”four times” option4=”one fourth” correct=”option2″]

This question was previously asked in

UPSC NDA-2 – 2020

Download PDF Attempt Online

The radiation force (F) on a totally reflecting surface is given by F = 2IA/c, where I is the intensity of light, A is the area of the surface, and c is the speed of light. The question states that the light has uniform intensity (I is constant) and impinges perpendicularly. When the area of the surface is halved, the new area A₂ = A₁/2. The new radiation force F₂ will be F₂ = 2IA₂/c = 2I(A₁/2)/c = (2IA₁/c) / 2 = F₁/2. Thus, the radiation force on it will become half.

Radiation force on a surface is proportional to the intensity of light and the area of the surface. For a totally reflecting surface, the force is twice that on a totally absorbing surface for the same intensity and area.

The radiation pressure is defined as the force per unit area (P = F/A). For a totally reflecting surface, the radiation pressure is P = 2I/c. The force is the product of pressure and area: F = P * A = (2I/c) * A.

More similar MCQ questions for Exam preparation:-