Physics · Wave Optics · NEET
Crossed polaroids P1 and P2 have axes at 90 degrees, so P2 blocks everything P1 passes (cos^2 90 = 0). When you put P3 in the middle at some angle theta, P3 changes the direction of vibration of the light. Now the light reaching P2 is no longer at 90 degrees to P2's axis. Its vibration has a component along P2's pass axis, so some light gets through. The middle sheet acts like a bridge: it rotates the plane of vibration in two steps instead of one big blocked step.
Read the question carefully. If I0 is the UNPOLARISED light falling on the first polaroid, then after P1 the intensity is I0/2 (a single polaroid always cuts unpolarised light to half). If the problem says I0 is already the intensity AFTER the first polaroid, then you start directly with I0. This one wording change decides whether the final answer is I0/8 or I0/8 with a different starting point. The NEET 2017 question gives unpolarised I0, so answer is I0/8. The NEET 2025 question gives I0 as the light after the first polaroid, and the answer is still I0/8 because the middle sheet is at 22.5 degrees, not 45.
Let the middle polaroid make angle theta with the first. The final light is I = (I0/2) cos^2(theta) cos^2(90 - theta) = (I0/2) cos^2(theta) sin^2(theta). Using cos(theta) sin(theta) = (1/2) sin(2theta), this becomes I = (I0/8) sin^2(2theta). This is maximum when sin(2theta) = 1, that is 2theta = 90, so theta = 45 degrees. Maximum output = I0/8 (taking I0 as the unpolarised input).
Since the outer two are crossed (90 degrees apart), and the middle one is at theta from the first, the middle one is at (90 - theta) from the last one. So you apply Malus law twice: first factor cos^2(theta) between P1 and P3, second factor cos^2(90 - theta) between P3 and P2. Students often wrongly use theta for both steps, which is the most common mistake in this problem.
The intensity drops straight back to zero. The middle sheet is the only reason any light passes. Take it out and you are back to two crossed polaroids with nothing to rotate the vibration, so P2 blocks everything. This is a favourite conceptual NEET question: adding a sheet increases the light, which feels wrong but is correct.
Two Polaroids P1 and P2 are placed with their axes perpendicular to each other. Unpolarised light I0 is incident on P1. A third polaroid P3 is kept in between P1 and P2 such that its axis makes an angle 45 degrees with that of P1. The intensity of transmitted light through P2 is
A polaroid sheet is placed between two crossed polaroids, with its polarization axis at 22.5 degrees from the polarization axis of one of the polaroids. The intensity of the transmitted light is (I0 is the intensity of polarized light after passing through the first polaroid)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
If I0 is the unpolarised input and the middle polaroid is at angle theta to the first, then final intensity I = (I0/2) cos^2(theta) cos^2(90 - theta) = (I0/8) sin^2(2theta). If I0 is already the light after the first polaroid, drop the 1/2 factor.
With unpolarised I0, the first polaroid already cuts it to I0/2. The best the two Malus steps can do together is another factor of 1/4 (at theta = 45), giving (I0/2)(1/4) = I0/8. You can never beat this with crossed outer polaroids.
For calculating the final intensity with crossed outer polaroids, the middle sheet at theta gives the same result whether you measure theta from P1 or P2, because sin^2(2theta) is symmetric about 45 degrees. But physically the middle sheet must sit BETWEEN the crossed pair, not outside, or no light comes out.
Use I = (I0/8) sin^2(2 x 30) = (I0/8) sin^2(60) = (I0/8)(3/4) = 3I0/32, taking I0 as unpolarised input.
Yes. Polaroids and Malus law are in NCERT Class 12 Wave Optics section 10.7, and the crossed-polaroids-with-a-third-sheet result has been asked directly in NEET 2017 and NEET 2025.
