JEE MAIN - Physics (2024 - 27th January Evening Shift - No. 16)
An object is placed in a medium of refractive index 3 . An electromagnetic wave of intensity $$6 \times 10^8 \mathrm{~W} / \mathrm{m}^2$$ falls normally on the object and it is absorbed completely. The radiation pressure on the object would be (speed of light in free space $$=3 \times 10^8 \mathrm{~m} / \mathrm{s}$$ ) :
$$6 \mathrm{~Nm}^{-2}$$
$$36 \mathrm{~Nm}^{-2}$$
$$18 \mathrm{~Nm}^{-2}$$
$$2 \mathrm{~Nm}^{-2}$$
Explanation
$$\begin{aligned}
& \text { Radiation pressure }=\frac{I}{\mathrm{~V}} \\
& =\frac{\mathrm{I} \cdot \mu}{\mathrm{c}} \\
& =\frac{6 \times 10^8 \times 3}{3 \times 10^8} \\
& =6 \mathrm{~N} / \mathrm{m}^2
\end{aligned}$$
Comments (0)
