Q.47. An unpolarized light of intensity I passes through polarizer and then through a certain optica

Q: What is the correct answer for JEE Main Physics Q47?

The correct answer is 30. When unpolarized light of intensity I passes through a polarizer, its intensity becomes $I_0 = I/2$. This polarized light then passes through an optically active solution, which rotates the plane of polarization by an angle $\phi$. Let the angle between the polarizer and analyser be $\theta$. If the light is rotated by $\phi$ by the solution, the angle between the transmitted light from the soluti

Q: Q.47. An unpolarized light of intensity I passes through polarizer and then through a certain optically active solution and finally it goes to analyser. If the angle between analyser and polariser is…

The correct answer is 30. When unpolarized light of intensity I passes through a polarizer, its intensity becomes $I_0 = I/2$. This polarized light then passes through an optically active solution, which rotates the plane of polarization by an angle $\phi$. Let the angle between the polarizer and analyser be $\theta$. If the light is rotated by $\phi$ by the solution, the angle between the transmitted light from the soluti

Question

Q.47. An unpolarized light of intensity I passes through polarizer and then through a certain optically active solution and finally it goes to analyser. If the angle between analyser and polariser is  and intensity of light emerged from analyser is  I, the angle of rotation of the light by the solution with respect to analyser is ________ degrees.

FirstInTest · Accepted Answer

When unpolarized light of intensity I passes through a polarizer, its intensity becomes $I_0 = I/2$. This polarized light then passes through an optically active solution, which rotates the plane of polarization by an angle $\phi$. Let the angle between the polarizer and analyser be $	heta$. If the light is rotated by $\phi$ by the solution, the angle between the transmitted light from the solution and the analyser's transmission axis becomes $(	heta - \phi)$ or $(\phi - 	heta)$. According to Malus's Law, the intensity emerging from the analyser is $I_{out} = I_0 \cos^2(	heta - \phi)$. Given $I_{out} = \frac{3}{8}I$ and $I_0 = I/2$. So, $\frac{3}{8}I = \frac{I}{2} \cos^2(	heta - \phi) \Rightarrow \frac{3}{4} = \cos^2(	heta - \phi)$. This means $\cos(	heta - \phi) = \pm \frac{\sqrt{3}}{2}$. So, $(	heta - \phi) = 30^\circ$ or $150^\circ$. The question implies that the angle between analyser and polarizer is $	heta$, and the light is rotated by the solution with respect to the analyser. This phrasing is a bit ambiguous. Let's assume the angle between the polarizer and analyser is $	heta$. The light from the polarizer has intensity $I_0 = I/2$. If the solution rotates the plane of polarization by an angle $\phi$, and the analyser is set at an angle $	heta$ with respect to the polarizer, then the angle between the light's polarization and the analyser's axis is $(	heta - \phi)$. So, $I_{final} = I_0 \cos^2(	heta - \phi) = (I/2) \cos^2(	heta - \phi)$. We are given $I_{final} = \frac{3}{8}I$. So, $\frac{3}{8}I = \frac{I}{2} \cos^2(	heta - \phi) \Rightarrow \frac{3}{4} = \cos^2(	heta - \phi)$. This implies $\cos(	heta - \phi) = \frac{\sqrt{3}}{2}$. Therefore, $(	heta - \phi) = 30^\circ$. The question asks for the angle of rotation of the light by the solution with respect to the analyser. This is $\phi$. However, the solution directly calculates $	heta = 30^\circ$ from $\cos	heta = \sqrt{3}/2$, implying that the angle of rotation is directly $	heta$. This suggests that the analyser is set at an angle $	heta$ relative to the *rotated* plane of polarization. If the polarizer is at $0^\circ$, the light after the polarizer is polarized at $0^\circ$. The solution rotates it by $\phi$. So it's polarized at $\phi$. The analyser is at an angle $	heta$ relative to the polarizer. So the angle between the light and analyser is $(	heta - \phi)$. The question states "angle between analyser and polariser is $	heta$". And "angle of rotation of the light by the solution with respect to analyser is ____ degrees". This is confusing. Let's follow the solution's interpretation: if the intensity after the polarizer is $I_0$, and the analyser is at angle $	heta$ to the polarizer, and the solution rotates the light by an angle $\phi$, then the final intensity is $I_0 \cos^2(	heta - \phi)$. If the question means that the angle $	heta$ is the angle between the polarizer and the *final* analyser setting, and the solution rotates the light such that the final intensity is $3/8 I$, then $(I/2) \cos^2(	heta) = 3/8 I \Rightarrow \cos^2	heta = 3/4 \Rightarrow \cos	heta = \sqrt{3}/2 \Rightarrow 	heta = 30^\circ$. This interpretation makes the question simpler and matches the solution's answer. The phrase "angle of rotation of the light by the solution with respect to analyser" is still problematic. It should be "angle between the analyser and the plane of polarization after passing through the solution". Assuming the solution's interpretation, the angle is $30^\circ$. Confidence lowered due to ambiguity.

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