Question 33.7: Calculating f for a converging lens. A convex meniscus lens ......
Calculating f for a converging lens. A convex meniscus lens (Figs. 33-2a and 33-17) is made from glass with n=1.50. The radius of curvature of the convex surface is 22.4 \mathrm{~cm} and that of the concave surface is 46.2 \mathrm{~cm}. (a) What is the focal length? (b) Where will the image be for an object 2.00 \mathrm{~m} away?
APPROACH We use Eq.33-4, noting that R_{2} is negative because it refers to the concave surface.
\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right) (33-4)
(a) R_{1}=22.4 \mathrm{~cm} and R_{2}=-46.2 \mathrm{~cm}.
Then
\begin{aligned}\frac{1}{f} & =(1.50-1.00)\left(\frac{1}{22.4 \mathrm{~cm}}-\frac{1}{46.2 \mathrm{~cm}}\right) \\& =0.0115 \mathrm{~cm}^{-1} .\end{aligned}
So
f=\frac{1}{0.0115 \mathrm{~cm}^{-1}}=87.0 \mathrm{~cm}
and the lens is converging. Notice that if we turn the lens around so that R_{1}=-46.2 \mathrm{~cm} and R_{2}=+22.4 \mathrm{~cm}, we get the same result.
(b) From the lens equation, with f=0.870 \mathrm{~m} and d_{\mathrm{o}}=2.00 \mathrm{~m}, we have
\begin{aligned}\frac{1}{d_{\mathrm{i}}}=\frac{1}{f}-\frac{1}{d_{\mathrm{o}}} & =\frac{1}{0.870 \mathrm{~m}}-\frac{1}{2.00 \mathrm{~m}} \\& =0.649 \mathrm{~m}^{-1}\end{aligned}
so d_{\mathrm{i}}=1 / 0.649 \mathrm{~m}^{-1}=1.54 \mathrm{~m}.