JEE MAIN - Physics (2023 - 1st February Morning Shift - No. 2)
'$$n$$' polarizing sheets are arranged such that each makes an angle $$45^{\circ}$$ with the preceeding sheet. An unpolarized light of intensity I is incident into this arrangement. The output intensity is found to be $$I / 64$$. The value of $$n$$ will be:
4
5
3
6
Explanation
After passing through first sheet
$$ I_1=\frac{I}{2} $$
After passing through second sheet
$$ I_2=I_1 \cos ^2\left(45^{\circ}\right)=\frac{I}{4} $$
After passing through $n^{\text {th }}$ sheet
$$ \begin{aligned} & I_{\mathrm{n}}=\frac{I}{2^{\mathrm{n}}}=\frac{I}{64} \\\\ & n=6 \end{aligned} $$
$$ I_1=\frac{I}{2} $$
After passing through second sheet
$$ I_2=I_1 \cos ^2\left(45^{\circ}\right)=\frac{I}{4} $$
After passing through $n^{\text {th }}$ sheet
$$ \begin{aligned} & I_{\mathrm{n}}=\frac{I}{2^{\mathrm{n}}}=\frac{I}{64} \\\\ & n=6 \end{aligned} $$
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