Consider a small light bulb,3 m to the left of a biconvex lens,which forms an image 2 m to the right of the lens.The lens is made of glass with index of refraction 1.4 and its two surfaces have equal radii of curvature.How large is the radius of curvature?
A)0.12 m
B)0.96 m
C)0.45 m
D)2 m
E)3 m

Question

Quizplus · Accepted Answer

The lens maker's formula for a lens in air is given by $\frac{1}{f} = (n-1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$, where $f$ is the focal length of the lens, $n$ is the refractive index of the lens material, and $R_1$ and $R_2$ are the radii of curvature of the two lens surfaces. For a biconvex lens with equal radii of curvature, $R_1 = R_2 = R$, and the formula simplifies to $\frac{1}{f} = (n-1) \left( \frac{2}{R} \right)$. Given that the object distance ($u$) is -3 m (negative because it's on the opposite side of the light's direction), and the image distance ($v$) is 2 m, we can use the lens equation $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ to find the focal length $f$. Substituting $u = -3$ m and $v = 2$ m gives $\frac{1}{f} = \frac{1}{2} - \frac{1}{-3} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$ m$^{-1}$, so $f = \frac{6}{5}$ m. Substituting $f = \frac{6}{5}$ m and $n = 1.4$ into the simplified lens maker's formula gives $\frac{5}{6} = (1.4 - 1) \left( \frac{2}{R} \right)$, solving for $R$ yields $R = 0.96$ m.

Consider a Small Light Bulb,3 M to the Left of a Biconvex