OPTION 1 (Alternative approach using relative refractive index)
Step 1: The focusing power of a lens depends on the relative refractive index of the lens material with respect to its surroundings. Write the lens maker relation as \(\frac{1}{f} = (\,{}^{a}n_{lens} - 1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\) in air, where \({}^{a}n_{lens}=n\).
Step 2: In a medium the relative index becomes \({}^{m}n_{lens} = \frac{n}{n_m}\). So \(\frac{1}{f_m} = \left(\frac{n}{n_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\).
Step 3: Taking the ratio of the two expressions removes the geometry term and gives \(f_m = \dfrac{(n-1)n_m}{n-n_m}\,f\).
Step 4: The medium here has index equal to \(n\). Then \({}^{m}n_{lens} = n/n = 1\), and \((\,{}^{m}n_{lens}-1)=0\), so \(\frac{1}{f_m}=0\). \[ \boxed{\,f_m = \infty\,} \] An index-matched lens is optically 'invisible' and acts as a plane slab.
OPTION 2 (Two-stage Malus argument)
Step 1: Polarized light has its field vector oscillating in one fixed plane; unpolarized light is a random mixture of all such planes. A polaroid transmits only the component of the field parallel to its pass axis.
Step 2: With only A and B present and their axes perpendicular, the field passed by A has zero component along B's axis, so \(I = I_0\cos^2 90^\circ = 0\) (dark).
Step 3: A middle polaroid C at angle \(\theta\) first projects the field onto its own axis ( factor \(\cos\theta\) ), then B projects that onto its axis ( factor \(\cos(90^\circ-\theta)=\sin\theta\) ). Intensity multiplies the squares: \(I_B = I_0\cos^2\theta\sin^2\theta\).
Step 4: This is non-zero except at \(\theta=0^\circ\) or \(90^\circ\); the once-blocked light is partly restored. \[ \boxed{\,\text{Yes, a third polaroid revives the light, peak } I_0/4 \text{ at } 45^\circ\,} \]