JEE Main 2024 · 27 January · evening · Single correct · +4 / −1
An object is placed in a medium of refractive index 3 . An electromagnetic wave of intensity 6 × 10⁸ W / m² falls normally on the object and it is absorbed completely. The radiation pressure on the object would be (speed of light in free space =3 × 10⁸ m / s ) :
- (a) 6 Nm⁻²
- (b) 36 Nm⁻²
- (c) 18 Nm⁻²
- (d) 2 Nm⁻²
Answer and solution
Answer (a) 6 Nm⁻²
Solution
Physics sir · 3 steps
Light slows down inside the medium, so divide the intensity by its speed there, v = c/n, not by c.
- Pressure from absorption. A wave that is fully absorbed hands over all its momentum. The pressure is the intensity divided by the speed of the wave where it hits the object.
- Speed in the medium. With refractive index n = 3, the wave travels at v = c/n = (3 × 10⁸)/3 = 1 × 10⁸ m s⁻¹.
P = I/v = nI/c = (3 × 6 × 10⁸)/(3 × 10⁸) = 6 N m⁻²
- So. The radiation pressure on the object is 6 N m⁻², which is option (a).
The idea to keep. For radiation pressure, use the wave's speed where it lands: P = I/v for full absorption and 2I/v for full reflection, with v = c/n inside a medium.
Answer from the NTA final answer key. Solution by PrepWiser.