Verified Solution • 2025–26 Session moderate

QJEEM23Ph24ea: JEE Main 2023 Physics - Class 12 JEEM23Ph24ea Solution: ' ' polarizing sheets are arranged such that each makes an…

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

Multiple Choice Options

  • A 4
  • B 5
  • C 3
  • D 6

Step-by-Step Verified Solution

After passing through first sheet

I1=I2I_1=\frac{I}{2}

After passing through second sheet

I2=I1cos⁡2(45∘)=I4I_2=I_1 \cos ^2\left(45^{\circ}\right)=\frac{I}{4}

After passing through nth n^{\text {th }} sheet

amp;In=I2n=I64amp;n=6\begin{aligned}& I_{\mathrm{n}}=\frac{I}{2^{\mathrm{n}}}=\frac{I}{64} \\\\& n=6\end{aligned}

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