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'n' polarizing sheets are arranged such that each makes an angle 45\degree 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:

Option: 1

4


Option: 2

3


Option: 3

5


Option: 4

6


Answers (1)

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According to Malus law:

\begin{aligned} & \mathrm{I}=\frac{\mathrm{I}_0}{2}\left[\cos ^2 45 \times \cos ^2 45 \times \cos ^2 45 \times \ldots(\mathrm{n}-1) \text { times }\right] \\ & \frac{\mathrm{I}_0}{64}=\frac{\mathrm{I}_0}{2} \times\left(\frac{1}{2}\right)^{\mathrm{n}-1} \\ & \frac{1}{32}=\frac{1}{2^{\mathrm{n}-1}} \Rightarrow \frac{1}{(2)^5}=\frac{1}{2^{\mathrm{n}-1}} \\ & \therefore \mathrm{n}-1=5 \\ & \therefore \mathrm{n}=6 \end{aligned}

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Deependra Verma

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