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JEE Main 2023 (1 Feb, Shift 1) · Polarisation

Polarisation question from JEE Main 2023 (1 Feb, Shift 1)

Q40JEE Main 2023 · 1 FEB · SHIFT 1Difficult

Polarisation

' $n$ ' polarizing sheets are arranged such that each makes an angle $45^\circ$ with the preceding 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:

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Answer: (d)

First sheet: $I/2$ ; each of the other $(n-1)$ sheets multiplies by $\cos^2 45^\circ = \frac{1}{2}$ .
$I_{out} = \frac{I}{2^n} = \frac{I}{64} \Rightarrow 2^n = 64$
So the answer is (d) $n = 6$ .

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