Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A . The intensity of light beyond B is found to be $\frac{\mathrm{I}}{2}$. Now another identical polarizer C is placed between A and B. The intensity beyond B is now found to be $\frac{\mathrm{I}}{8}$. The angle between polarizer A and C is :

JEE Main · 2018 · 08 Apr 2018 · Offline · Wave Optics

Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A . The intensity of light beyond B is found to be

\frac{I}{2}

. Now another identical polarizer C is placed between A and B. The intensity beyond B is now found to be

\frac{I}{8}

. The angle between polarizer A and C is :

(A) $6 0^{\circ}$
(B) $0^{\circ}$
(C) $3 0^{\circ}$
(D) $4 5^{\circ}$ ✓ Correct

Answer: (D)

Solution

Through the first polariser the unpolarised light drops to

\frac{I}{2}

, and with A and B parallel this passes on unchanged. Inserting C at angle

θ

to A, Malus's law is applied twice: after C the intensity is

\frac{I}{2} cos^{2} θ

, and after B, now

θ

beyond C, it is

\frac{I}{2} cos^{2} θ cos^{2} θ = \frac{I}{2} cos^{4} θ

. Setting this equal to

\frac{I}{8}

gives

cos^{4} θ = \frac{1}{4}

, so

cos^{2} θ = \frac{1}{2}

and

θ = 4 5^{\circ}

JEE Main · 2018 · 08 Apr 2018 · Offline · Single correct option

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JEE Main · 2018 · 08 Apr 2018 · Offline · Wave Optics

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