Question

Figure shows a simple version of a zoom lens. The converging lens has focal length $f_1$ and the diverging lens has focal length $f_2=-|f_2|$ .The two lenses are separated by a variable distance $d$ that is always less than $f_1$ . Also, the magnitude of the focal length of the diverging lens satisfies the inequality $|f_2|>(f_1-d)$ .To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius $r_0$ entering the converging lens. It can be shown that the radius of the ray bundle decreases to $r_0^{\prime }=\frac{r_0^{\prime }\left(f_{-}d\right)}{f_1}$ at the point that it enters the diverging lens . Final image $I^{\prime }$ is formed a distance $S_2^{\prime }=\frac{|f_2|(f_1-d)}{(|f_2|-f_1+d)}$ the right of the diverging lens. If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left of the diverging lens; they will eventually expand to the original radius $r_0$ at same point Q. The distance from the final image $I^{\prime }$ to the point Q is the effective focal length $f$ of the lens combination; if the combination were replaced by a single lens of focal length f placed at Q, parallel rays would still be brought to a focus at $I^{\prime }$ . The effective focal length is given by $f=\frac{f_1|f_2|}{(|f_2|-f_1+d)}$
Assume that $f_1=12.0cm,f_2=-18.0cm,$ and the separation $d$ is adjustable $O$ and $3.0cm$
The final image distance $S_2^{\prime }$ will be such that :

• A. $0\le S_2^{\prime }\le |f_2|$
• B. $|f_2|\le S_2^{\prime }\le 2|f_2|$
• C. $2|f_2|\le S_2^{\prime }\le \infty$
• D. Any of the above three options

Answer 1

The correct option is D Any of the above three options

$S_2^{\prime }$ is given by formula

$S_2^{\prime }=\frac{|f_2|(f_1-d)}{(|f_2|-f_1+d)}$

$f_1=12.0cm$, $f_2=-18.0cm$, $d$ is varying between $0\to 3cm$

let $d=3cm$

$S_2^{\prime }=\frac{|-18|(12-3)}{(|-18|-12+3)}$

$S_2^{\prime }=\frac{18\times 9}{(18-12+3)}$

$S_2^{\prime }=\frac{18\times 9}{9}$

$S_2^{\prime }=18.0cm$ and $f_2=18.0cm$

now for $d=0cm$

$S_2^{\prime }=\frac{|-18|(12-0)}{(|-18|-12+0)}$

$S_2^{\prime }=\frac{18\times 12}{(18-12+0)}$

$S_2^{\prime }=\frac{18\times 12}{6}$

$S_2^{\prime }=36.0cm$

from above calculations $18.0\le S_2^{\prime }\le 36.0$

for boundary cases $S_2=|f_2|$ or $S_2=2|f_2|$

so any of the three condition is possible.

option D is correct.