Question 33.16: Microscope. A compound microscope consists of a 10× eyepiece......
Microscope. A compound microscope consists of a 10 \times eyepiece and a 50 \times objective 17.0 \mathrm{~cm} apart. Determine (a) the overall magnification, (b) the focal length of each lens, and (c) the position of the object when the final image is in focus with the eye relaxed. Assume a normal eye, so N=25 \mathrm{~cm}.
APPROACH The overall magnification is the product of the eyepiece magnification and the objective magnification. The focal length of the eyepiece is found from Eq. 33-6a or 33-9 for the magnification of a simple magnifier. For the objective lens, it is easier to next find d_{\mathrm{o}} (part c ) using Eq. 33-8 before we find f_{\mathrm{o}}.
M=\frac{\theta^{\prime}}{\theta}=\frac{h / f}{h / N}=\frac{N}{f} \cdot\left[\begin{array}{c}\text { eye focused at } \infty ; \\N=25 \mathrm{~cm} \text { for normal eye }\end{array}\right] (33-6a)
M_{\mathrm{e}}=\frac{N}{f_{\mathrm{e}}} (33-9)
m_{\mathrm{o}}=\frac{h_{\mathrm{i}}}{h_{\mathrm{o}}}=\frac{d_{\mathrm{i}}}{d_{\mathrm{o}}}=\frac{\ell-f_{\mathrm{e}}}{d_{\mathrm{o}}} (33-8)
(a) The overall magnification is (10 \times)(50 \times)=500 \times.
(b) The eyepiece focal length is (Eq. 33-9) f_{\mathrm{e}}=N / M_{\mathrm{e}}=25 \mathrm{~cm} / 10=2.5 \mathrm{~cm}. Next we solve Eq. 33-8 for d_{0}, and find
d_{\mathrm{o}}=\frac{\ell-f_{\mathrm{e}}}{m_{\mathrm{o}}}=\frac{(17.0 \mathrm{~cm}-2.5 \mathrm{~cm})}{50}=0.29 \mathrm{~cm}
Then, from the thin lens equation for the objective with d_{\mathrm{i}}=\ell-f_{\mathrm{e}}=14.5 \mathrm{~cm} (see Fig. 33-40a),
\frac{1}{f_{\mathrm{o}}}=\frac{1}{d_{\mathrm{o}}}+\frac{1}{d_{\mathrm{i}}}=\frac{1}{0.29 \mathrm{~cm}}+\frac{1}{14.5 \mathrm{~cm}}=3.52 \mathrm{~cm}^{-1}
so f_{\mathrm{o}}=1 /\left(3.52 \mathrm{~cm}^{-1}\right)=0.28 \mathrm{~cm}.
(c) We just calculated d_{\mathrm{o}}=0.29 \mathrm{~cm}, which is very close to f_{\mathrm{o}}.
