Let \( I_0 \) be the intensity of unpolarized light incident on the first polaroid.
- The transmitted light intensity after passing through the first polaroid is:
\(I_1 = \frac{I_0}{2}.\)
- The intensity after passing through the second polaroid at an angle \( \theta \) is:
\(I_2 = I_1 \cos^2 \theta = \frac{I_0}{2} \cos^2 \theta.\)
- Similarly, for the third polaroid rotated by \( 90^\circ - \theta \), the transmitted intensity is:
\(I_3 = I_2 \cos^2 (90^\circ - \theta) = \frac{I_0}{2} \cos^2 \theta \sin^2 \theta.\)
- The intensity will be maximum when \( \sin 2\theta = 1 \), which occurs at \( \theta = 45^\circ \).
The Correct answer is: \(45^\circ\)
Calculate the angle of minimum deviation of an equilateral prism. The refractive index of the prism is \(\sqrt{3}\). Calculate the angle of incidence for this case of minimum deviation also.