JEECUP (UP Polytechnic) Malus' Law — practice questions

Explore 2 free Malus' Law questions in JEECUP (UP Polytechnic) Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Malus' Law questions and exercises with answers

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**Malus' law** gives the transmitted intensity through a second polarizer as:Read the answer and worked solution →Light of intensity $I_0$ passes through two polarizers whose axes make 60° with each other. The transmitted intensity is:Read the answer and worked solution →

Worked practice questions

Malus' law gives the transmitted intensity through a second polarizer as:

$I = I_0 \sin\theta$
$I = I_0 \cos^2\theta$
$I = I_0 \sin^2\theta$
$I = I_0$

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Correct answer: B. $I = I_0 \cos^2\theta$

When plane-polarized light (intensity $I_0$) hits a polarizer whose axis makes angle $\theta$ with its polarization, the transmitted intensity is $I_0 \cos^2\theta$. At $\theta = 0$, full intensity at $\theta = 90°$, zero (crossed polarizers).

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Light of intensity $I_0$ passes through two polarizers whose axes make 60° with each other. The transmitted intensity is:

$I_0/8$
$I_0/4$
$I_0/2$
$I_0 \cos^2 60°$

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Correct answer: A. $I_0/8$

**First polarizer:** $I_0 \to I_0/2$ (unpolarized). **Second polarizer (Malus):** $I_0/2 \times \cos^2 60° = I_0/2 \times 1/4 = I_0/8$.