JEE Physics 2023 Question with Solution

Given: Unpolarized light of intensity $I$ is incident on $n$ polarizing sheets, and each successive sheet makes an angle $45^\circ$ with the preceding sheet.

Find: The value of $n$ when the final intensity is $\frac{I}{64}$.

After passing through the first sheet,

$$I_1 = \frac{I}{2}$$

After passing through the second sheet,

$$I_2 = I_1 \cos^2(45^\circ) = \frac{I}{4}$$

After passing through $n$th sheet,

$$I_n = \frac{I}{2^n} = \frac{I}{64}$$

So,

$$2^n = 64 = 2^6$$

Hence,

$$n = 6$$

Therefore, the computed value is $6$. The solution states the correct option is A, but this conflicts with the listed options; among the given options, $6$ corresponds to option B.

Given: Each polarizer is inclined by $45^\circ$ relative to the previous one.

Find: Number of sheets required so that output intensity becomes $\frac{I}{64}$.

For unpolarized light, the first polarizer transmits half the intensity:

$$I_1 = \frac{I}{2}$$

For every next sheet at angle $45^\circ$,

$$\cos^2 45^\circ = \frac{1}{2}$$

So each additional sheet again multiplies intensity by $\frac{1}{2}$. Therefore after $n$ sheets,

$$I_n = \frac{I}{2^n}$$

Given,

$$\frac{I}{2^n} = \frac{I}{64}$$

Cancelling $I$,

$$2^n = 64$$

Now,

$$64 = 2^6$$

Thus,

$$n = 6$$

Therefore, the correct option from the given options is B.