A linearly polarized monochromatic light of intensity 10 lumen is incident on a polarizer. The angle between t

Question

A linearly polarized monochromatic light of intensity 10 lumen is incident on a polarizer. The angle between the direction of polarization of the light and that of the polarizer such that the intensity of output light is 2.5 lumen is:

Accepted Answer

1. **Identify Given Values:**
   - Incident Intensity ($I_0$) = 10 lumen (Correcting typo '101010').
   - Transmitted/Output Intensity ($I$) = 2.5 lumen (Correcting typo '2.52.52.5').
2. **Formula:** According to **Malus' Law**, when linearly polarized light is incident on a polarizer, the transmitted intensity is given by:
   $$ I = I_0 \cos^2 	heta $$
   where $	heta$ is the angle between the direction of polarization of the incident light and the pass axis of the polarizer.
3. **Calculation:**
   $$ 2.5 = 10 \cos^2 	heta $$
   $$ \cos^2 	heta = \frac{2.5}{10} = \frac{1}{4} = 0.25 $$
   Taking the square root:
   $$ \cos 	heta = 0.5 $$
   $$ 	heta = \cos^{-1}(0.5) = 60^\circ $$
4. **Conclusion:** The angle must be $60^\circ$.

Step-by-Step Solution

Practice All PHYSICS Questions