Physics Q47 – JEE-MAIN 2026

An unpolarized light of intensity

I_{0}

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

0^{\circ}

and intensity of light emerged from analyser is

\frac{3}{8} I_{0}

, the angle of rotation of the light by the solution with respect to analyser is _______ degrees.

Enter Calculated Numerical Value:

Clue 1 of 3: Polarization of Light

Understand how unpolarized light changes after passing through a polarizer and how an optically active solution affects polarized light.

🧠 Full Solution Path

Step 1: Intensity after Polarizer✦ Active

When unpolarized light of intensity $I_{0}$ passes through a polarizer, the intensity of the transmitted plane-polarized light becomes half of the incident intensity.

I_{polarized} = \frac{I_{0}}{2}

Step 2: Apply Malus's Law○ Expand

The optically active solution rotates the plane of polarization by an angle, let's call it $α$ . Since the analyser is initially aligned with the polarizer (angle $0^{\circ}$ ), the angle between the plane of polarization of the light (after the solution) and the analyser's transmission axis is $α$ . According to Malus's Law, the intensity of light emerged from the analyser is:

I_{analyser} = I_{polarized} \cos^{2} α

Substitute $I_{polarized} = \frac{I_{0}}{2}$ :

I_{analyser} = \frac{I_{0}}{2} \cos^{2} α

💡 Teacher's Secret Hint

Ensure the angle used in Malus's Law is the angle between the incident polarization plane and the analyser's axis.

Step 3: Solve for the Angle of Rotation○ Expand

We are given that the intensity of light emerged from the analyser is $\frac{3}{8} I_{0}$ . Equating this to the expression from Malus's Law:

\frac{3}{8} I_{0} = \frac{I_{0}}{2} \cos^{2} α

Divide both sides by $I_{0}$ and simplify:

\frac{3}{8} = \frac{1}{2} \cos^{2} α

\cos^{2} α = \frac{3}{8} \times 2 = \frac{3}{4}

\cos α = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}

The angle of rotation is typically taken as the smallest positive angle. Therefore, $α = 30^{\circ}$ .

💡 Teacher's Secret Hint

Remember that $\cos^{2} α = \frac{3}{4}$ implies $\cos α = \pm \frac{\sqrt{3}}{2}$ . Consider the principal value for the angle of rotation.

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Physics Question 47 – JEE-MAIN 2026

Clue 1 of 3: Polarization of Light

🧠 Full Solution Path