Question

The light from a torch is normally incident on the surface at intensity of illumination $I_1$. If the surface is tilted to ${45}^{\circ }$ from initial position, then what will be intensity of illumination $I_2$ on the surface in terms of $I_1$ ?

• A. $I_2=\frac{I_1}{\sqrt{2}}$
• B. $I_2=I_1$
• C. $I_2=2I_1$
• D. $I_2=4I_1$

Answer 1

The correct option is A $I_2=\frac{I_1}{\sqrt{2}}$


Intensity $I=\frac{P}{A}$ where $P$ is the power of incident light and $A$ is the area of surface.

Then, $I_1=\frac{P}{A}$
and $I_2=\frac{P}{A^{\prime }}$
From the diagram, $A^{\prime }\cos {45}^{\circ }=A$
$\Rightarrow I_2=\frac{P}{\frac{A}{\cos {45}^{\circ }}}=\frac{I_1}{\sqrt{2}}$