Question

# Figure shows a simple version of a zoom lens. TheÂ converging lens has focal length $$f_1$$Â and the diverging lensÂ has focal length $$f_2 = - |f_2|$$ .The two lenses are separated byÂ a variable distance $$d$$ that is always less than $$f_1$$ . Also, theÂ magnitude of the focal length of the diverging lens satisfiesÂ the inequality $$|f_2| > (f_1 - d)$$ .To determine the effectiveÂ focal length of the combination lens, consider a bundle ofÂ parallel rays of radius $$r_0$$ entering the converging lens. It canÂ be shown that the radius of the ray bundle decreases toÂ  $${r}'_0 = \dfrac{{r}'_0 (f_ - d)}{f_1}$$Â at the point that it enters the diverging lens . Final image $$I'$$Â is formed a distance $${S}'_2 = \dfrac{|f_2| (f_1 - d)}{(|f_2| - f_1 + d)}$$Â the right of the diverging lens. If the rays that emerge fromÂ the diverging lens and reach the final image point areÂ extended backward to the left of the diverging lens; theyÂ will eventually expand to the original radius $$Â r_0$$Â at sameÂ point Q. The distance from the final image $$Â I'$$Â to the point QÂ is the effective focal length $$Â f$$Â of the lens combination; if theÂ combination were replaced by a single lens of focal length fÂ placed at Q, parallel rays would still be brought to a focus at $$Â I'$$ .Â Â The effective focal length is given by $$f = \dfrac{f_1|f_2|}{(|f_2| - f_1 + d)}$$Â Â Assume that $$f_1 = 12.0\,cm, f_2 = -18.0\,cm,$$Â  and theÂ separation $$d$$ is adjustable $$O$$ and $$3.0 \,cm$$The final image distance $${S}'_2$$ will be such that :Â

A
0S2|f2|
B
|f2|S22|f2|
C
2|f2|S2
D
Any of the above three options

Solution

## The correct option is D Any of the above three options$$S'_2$$ is given by formula $${S}'_2 = \dfrac{|f_2| (f_1 - d)}{(|f_2| - f_1 + d)}$$$$f_1=12.0\,cm$$, $$f_2=-18.0\,cm$$, $$d$$ is varying between $$0\to 3\,cm$$let $$d=3\,cm$$$${S}'_2 = \dfrac{|-18| (12 - 3)}{(|-18| - 12 + 3)}$$$${S}'_2 = \dfrac{18\times9}{(18 - 12 + 3)}$$$${S}'_2 = \dfrac{18\times9}{9}$$$${S}'_2 = 18.0\,cm$$ and $$f_2=18.0\,cm$$now for  $$d=0\,cm$$$${S}'_2 = \dfrac{|-18| (12 - 0)}{(|-18| - 12 + 0)}$$$${S}'_2 = \dfrac{18\times12}{(18 - 12 + 0)}$$$${S}'_2 = \dfrac{18\times12}{6}$$$${S}'_2 = 36.0\,cm$$from above calculations $$18.0 \leq {S}'_2 \leq 36.0$$ 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.Physics

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