Question

FIGURE $CP35.50$shows a simple zoom lens in which the magnitudes of both focal lengths are $f$. If the spacing $d<f$, the image of the converging lens falls on the right side of the diverging lens. Our procedure of letting the image of the first lens act as the object of the second lens will continue to work in this case if we use a negative object distance for the second lens. This is called a virtual object. Consider an object very far to the left $(s\approx \infty )$of the converging lens. Define the effective focal length as the distance from the midpoint between the lenses to the final image.

a. Show that the effective focal length is

$f_{eff}=\frac{f^2-fd+{\displaystyle \frac{1}{2}}{d}^2}{d}$

b. What is the zoom for a lens that can be adjusted from $d=\frac{1}{2}f$to $d=\frac{1}{4}f$?

Short answer

a. The required statement is proved below.

b. The zoom fir lens is zoom= $2.5$.

Step-by-step solution

Part (a) step 1: Given information

We need to show the effective focal length is $f_{eff}=\frac{f^2-fd+{\displaystyle \frac{1}{2}}{d}^2}{d}$

Part (a) step 2: Calculation

To prove the formula

$f_{eff}=\frac{f^2-fd+{\displaystyle \frac{1}{2}}{d}^2}{d}$

Now,

$$\begin{gathered} \frac{1}{d-f}+\frac{1}{s_2^{\text{'}}}=\frac{1}{-f} \\ {s}_2^{\text{'}}=\frac{-f(d-f)}{(d-f)-(-f)} \\ {s}_2^{\text{'}}=\frac{f_2-fd}{d} \\ {f}_{eff}=s_2^{\text{'}}+\frac{1}{2}d \\ {f}_{eff}=\frac{f^2-fd}{d}+\frac{1}{2}d \\ {f}_{eff}=\frac{f_2-fd}{d}+\frac{{\displaystyle \frac{1}{2}}{d}^2}{d} \\ {f}_{eff}=\frac{f^2-fd+{\displaystyle \frac{1}{2}}{d}^2}{d} \end{gathered}$$

Here $f_{eff}$is effective focal length.

Part (b) step 1: Given information

We need to find the zoom for a lens that can be adjusted from $d=\frac{1}{2}f$to $d=\frac{1}{4}f$.

Part (b) step 2: Calculation

Calculation for finding zoom

zoom=$\frac{{\left(f_{eff}\right)}_{d=1/2f}}{{\left(f_{eff}\right)}_{d=1/2f}}$

zoom=$\frac{{\displaystyle \frac{f^2-f\left({\displaystyle \frac{1}{2}}f\right)+{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{1}{4}}f\right)}^2}{{\displaystyle \frac{1}{4}}f}}}{{\displaystyle \frac{f^2-f\left({\displaystyle \frac{1}{2}}f\right)+{\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{1}{2}}f\right)}^2}{{\displaystyle \frac{1}{2}}f}}}$

zoom=$\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{4}}}\frac{f^2-{\displaystyle \frac{1}{4}}{f}^2+{\displaystyle \frac{1}{32}}{f}^2}{f^2-{\displaystyle \frac{1}{2}}{f}^2+{\displaystyle \frac{1}{8}}{f}^2}$

zoom=$\frac{4}{2}\frac{1-{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{32}}}{1-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{8}}}$

zoom=$\left(2\right)\frac{{\displaystyle \frac{25}{32}}}{{\displaystyle \frac{5}{8}}}-\left(2\right)\frac{5}{4}$

zoom=$\frac{5}{2}$

zoom=$2.5$