Q.42. An unpolarized light of certain intensity passes through a combination of two polarizers whose transmission axes are at and , respectively, with respect to the horizontal axis. A third polarize

Question

Q.42. An unpolarized light of certain intensity passes through a combination of two polarizers whose transmission axes are at  and , respectively, with respect to the horizontal axis. A third polarizer with its transmission axis at  with the horizontal axis is placed between the two existing polarizers. The ratio of the output intensities with and without the third polarizer is

FirstInTest · Accepted Answer

Correct answer: C. Sol. Let the initial intensity of unpolarized light be $I_0$. Case 1: Without the third polarizer. Light passes through polarizer 1 (axis at $30^\circ$) and then polarizer 2 (axis at $90^\circ$). Intensity after polarizer 1: $I_1 = I_0/2$. Intensity after polarizer 2: $I_2 = I_1 \cos^2(90^\circ - 30^\circ) = (I_0/2) \cos^2(60^\circ) = (I_0/2) (1/2)^2 = I_0/8$. Case 2: With the third polarizer. Light passes through polarizer 1 (axis at $30^\circ$), then polarizer 3 (axis at $60^\circ$), then polarizer 2 (axis at $90^\circ$). Intensity after polarizer 1: $I_1 = I_0/2$. Intensity after polarizer 3: $I_3 = I_1 \cos^2(60^\circ - 30^\circ) = (I_0/2) \cos^2(30^\circ) = (I_0/2) (\sqrt{3}/2)^2 = (I_0/2) (3/4) = 3I_0/8$. Intensity after polarizer 2: $I_4 = I_3 \cos^2(90^\circ - 60^\circ) = (3I_0/8) \cos^2(30^\circ) = (3I_0/8) (\sqrt{3}/2)^2 = (3I_0/8) (3/4) = 9I_0/32$. Ratio of output intensities (with / without third polarizer) = $\frac{I_4}{I_2} = \frac{9I_0/32}{I_0/8} = \frac{9}{32} 	imes 8 = \frac{9}{4}$. Therefore, option (3) is correct.

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