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Standard XII

Physics

Cutting of Thin Lenses

Fig 28.86 sho...

Question

Fig 28.86 shows a simple version of zoom lens. The converging lens has focal length $f_{1}$ and 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 $| f_{2} | > f_{1} - d_{0}$ . To find the focal length of the combination lens consider a bundle of parallel rays of radius $r_{0}$ entering the converging lens.
1. Find the radius of the ray bundle $r_{2}$ at the entrance of diverging lens.

r_{2} = \frac{r_{1} f_{1}}{f_{1} + d}

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r_{2} = \frac{r_{1} (f_{1} - d)}{f_{1}}

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r_{2} = \frac{r_{1} d}{f_{1}}

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r_{2} = \frac{r_{1} d}{f_{1} + d}

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Solution

The correct option is A $r_{2} = \frac{r_{1} (f_{1} - d)}{f_{1}}$
$\frac{r_{1}}{f_{1}} = \frac{r_{1} - r_{2}}{d} = tan θ$
$\frac{r_{2}}{d} = \frac{r_{1} (f_{1} - d)}{f_{1} d}$
$o r r_{2} = \frac{r_{1} (f_{1} - d)}{f_{1}}$

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The correct option is A r2=r1(f1−d)f1r1f1=r1−r2d=tanθr2d=r1(f1−d)f1dorr2=r1(f1−d)f1

The correct option is A $r_{2} = \frac{r_{1} (f_{1} - d)}{f_{1}}$
$\frac{r_{1}}{f_{1}} = \frac{r_{1} - r_{2}}{d} = tan θ$
$\frac{r_{2}}{d} = \frac{r_{1} (f_{1} - d)}{f_{1} d}$
$o r r_{2} = \frac{r_{1} (f_{1} - d)}{f_{1}}$