JEE Main 2026 — Physics Optics

Let the intensity of the incident unpolarized light be $I_0$I0​.

After passing through the first polarizer, the intensity becomes $I_1 = \dfrac{I_0}{2}$I1​=2I0​​.

Without the third polarizer, the light passes directly from the first polarizer (at $30^{\circ}$30∘) to the second polarizer (at $90^{\circ}$90∘). The angle between their transmission axes is $90^{\circ} - 30^{\circ} = 60^{\circ}$90∘−30∘=60∘.

Using Malus's law, the output intensity without the third polarizer is:

$$I_{without} = I_1 \cos^2(60^{\circ}) = \dfrac{I_0}{2} \times \left(\dfrac{1}{2}\right)^2 = \dfrac{I_0}{8}$$Iwithout​=I1​cos2(60∘)=2I0​​×(21​)2=8I0​​

When the third polarizer (at $60^{\circ}$60∘) is placed between them, the light first passes from the first polarizer to the third polarizer. The angle between their axes is $60^{\circ} - 30^{\circ} = 30^{\circ}$60∘−30∘=30∘.

The intensity after the third polarizer is:

$$I_3 = I_1 \cos^2(30^{\circ}) = \dfrac{I_0}{2} \times \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3I_0}{8}$$I3​=I1​cos2(30∘)=2I0​​×(23​​)2=83I0​​

The light then passes from the third polarizer to the second polarizer. The angle between their axes is $90^{\circ} - 60^{\circ} = 30^{\circ}$90∘−60∘=30∘.

The final output intensity with the third polarizer is:

$$I_{with} = I_3 \cos^2(30^{\circ}) = \dfrac{3I_0}{8} \times \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{9I_0}{32}$$Iwith​=I3​cos2(30∘)=83I0​​×(23​​)2=329I0​​

The ratio of the output intensities with and without the third polarizer is:

$$\text{Ratio} = \dfrac{I_{with}}{I_{without}} = \dfrac{\dfrac{9I_0}{32}}{\dfrac{I_0}{8}} = \dfrac{9}{4}$$Ratio=Iwithout​Iwith​​=8I0​​329I0​​​=49​

Answer: $9/4$9/4