JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Wave Optics

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n

' polarizing sheets are arranged such that each makes an angle

4 5^{\circ}

with the proceeding sheet. An unpolarized light of intensity

I

is incident into this arrangement. The output intensity is found to be

\frac{I}{64}

. The value of

n

will be:

(A)3
(B)6✓ Correct
(C)5
(D)4

Answer: (B)

Solution

The first sheet, receiving unpolarised light, passes half the intensity,

\frac{I}{2}

. Each later sheet is turned

4 5^{\circ}

from the one before, so by Malus's law it transmits a fraction

cos^{2} 4 5^{\circ} = \frac{1}{2}

. After all

n

sheets the emerging intensity is

I_{n} = \frac{I}{2} (\frac{1}{2})^{n - 1} = \frac{I}{2 ^{n}} .

Setting this equal to

\frac{I}{64}

gives

2^{n} = 64 = 2^{6}

, so

n = 6

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Single correct option

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