A double-convex thin glass lens has equal-magnitude radii of curvature. The focal length of this lens is $+ 31.2 \mathrm {~cm}$ and the index of refraction of the glass is $1.52$. The lens is to be replaced with a plano-convex glass lens of the same focal length and thickness. The radius of curvature of the convex surface of the replacement lens is $13.0 \mathrm {~cm}$. What should be the index of refraction of the glass of the plano-convex lens?
A) $1.40$
B) $1.42$
C) $1.38$
D) $1.36$
E) $1.44$

Question

Quizplus · Accepted Answer

The focal length $f$ of a lens in air is given by the lensmaker's formula: $$ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) $$ where $n$ is the index of refraction of the lens material, and $R_1$ and $R_2$ are the radii of curvature of the two lens surfaces. For a double-convex lens with equal radii of curvature, $R_1 = -R_2$, and for a plano-convex lens, $R_2 = \infty$.Given that the double-convex lens and the plano-convex lens have the same focal length, we can set their $1/f$ values equal to each other to find the index of refraction for the plano-convex lens.For the double-convex lens, $n = 1.52$ and $f = +31.2$ cm. For the plano-convex lens, $R_1 = 13.0$ cm and $R_2 = \infty$, so $\frac{1}{R_2} = 0$.Setting the two $1/f$ values equal and solving for the index of refraction of the plano-convex lens gives us $n = 1.42$.

A Double-Convex Thin Glass Lens Has Equal-Magnitude Radii of Curvature +31.2 cm+ 31.2 \mathrm {~cm}+31.2 cm

A Double-Convex Thin Glass Lens Has Equal-Magnitude Radii of Curvature $+ 31.2 \mathrm {~cm}$