'n' polarizing sheets are arranged such that each makes an angle 45° with the preceeding sheet. An unpolarized light of intensity I is incident into this arrangement. The output intensity is found to be I / 64. The value of n will be:

Solution

Physics sir · 4 steps

The first sheet halves the light, and every sheet after it halves it again, so count the halvings.

1. First sheet. Unpolarised light passing through a polariser keeps half its intensity, whatever the sheet's orientation. After sheet 1 the intensity is I/2.
2. Every later sheet. The light reaching each later sheet is already polarised. By Malus's law it is multiplied by cos² 45° = 1/2. There are n - 1 such sheets.

   Iₒᵤₜ = I/2(1/2)ⁿ⁻¹ = I/2ⁿ

3. Match the output. Set I/2ⁿ = I/64. Since 64 = 2⁶, this gives n = 6.
4. So. There are 6 polarising sheets, which is option (d).

The idea to keep. Unpolarised light loses half at the first polariser. Malus's law cos² θ applies only from the second sheet onwards, because only then is the incoming light polarised.

Answer from the NTA final answer key. Solution by PrepWiser.