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Step-by-Step Solution
Step 1: Understand the situation
Unpolarized light of intensity I falls on two Polaroids, P_1 and P_2 , which together initially reduce the intensity to I/2 . Then, by rotating P_2 through some angle \phi , the intensity further changes to 3I/8 . We need to find the angle \phi through which P_2 should be rotated.
Step 2: Recall Malus's law
When unpolarized light of intensity I_0 passes through the first Polaroid, its intensity becomes I_0/2 . If the second Polaroid is placed at an angle \theta relative to the axis of the first, the transmitted intensity I_{\text{out}} is given by:
I_{\text{out}} = \frac{I_0}{2} \cos^2 \theta
Step 3: Match initial condition ( I/2 on the screen)
The problem states that, with P_1 and P_2 placed initially, the observed intensity is I/2 (compared to the original I ). This configuration fixes the relative alignment of P_1 and P_2 such that the transmitted intensity from the two-Polaroid system is I/2 before we rotate P_2 .
Step 4: Apply Malus's law for the new intensity ( 3I/8 )
After rotating P_2 by an angle \phi from its initial orientation, Malus's law tells us the new transmitted intensity is:
\frac{I_0}{2} \cos^2 \phi
According to the problem, this becomes 3I/8 . Therefore, we set up the equation:
\frac{I_0}{2} \cos^2 \phi = \frac{3I}{8}.
Step 5: Solve for \cos^2 \phi
We note that initially, I_0 = I when no Polaroids are in place. Substituting I_0 = I in the above relation gives:
\frac{I}{2} \cos^2 \phi = \frac{3I}{8}
\quad \Longrightarrow \quad
\cos^2 \phi = \frac{3}{4}.
Step 6: Find the angle \phi
Taking the positive square root for the physically relevant angle, we get:
\cos \phi = \frac{\sqrt{3}}{2}
\quad \Longrightarrow \quad
\phi = 30^\circ.
Step 7: Final Answer
Hence, the Polaroid P_2 should be rotated by an angle of 30^\circ .
