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(B) For a thick lens,the focal length $f$ is given by the lens maker's formula for thick lenses: $\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R_1} - \frac

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$x < y$

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(B) For a thick lens,the focal length $f$ is given by the lens maker's formula for thick lenses: $\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} + \frac{(\mu - 1)t}{\mu R_1 R_2} \right]$.Case $1$: Rays incident on the plane surface $(R_1 = \infty, R_2 = -R)$.$\frac{1}{f} = (\mu - 1) \left[ 0 - \frac{1}{-R} + 0 \right] = \frac{\mu - 1}{R}$.Thus,$f = \frac{R}{\mu - 1}$. The distance $x$ is measured from the plane surface,so $x = f = \frac{R}{\mu - 1}$.Case $2$: Rays incident on the curved surface $(R_1 = R, R_2 = \infty)$.$\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R} - 0 + 0 \right] = \frac{\mu - 1}{R}$.Here,the focal length is the same,but the refraction occurs at the curved surface first. The rays converge at a distance $y$ from the curved surface. Due to the thickness $t$ of the lens,the effective distance $y$ from the curved surface is $y = f + \frac{t}{\mu} = \frac{R}{\mu - 1} + \frac{t}{\mu}$.Comparing $x$ and $y$,we see that $y = x + \frac{t}{\mu}$,which implies $y > x$ or $x < y$.

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