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  4. A double convex lens made of glass (refractive index n =1.5) has both radii of curvature of magnitude 20 cm. Incident light rays parallel to the axis of the lens will converge at a distance L such that

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Q. A double convex lens made of glass (refractive index $n =1.5$) has both radii of curvature of magnitude $20 \,cm$. Incident light rays parallel to the axis of the lens will converge at a distance $L$ such that

AFMCAFMC 2010Ray Optics and Optical Instruments

Solution:

Here, $n=1.5$ as per sign convention followed $R_{1}=+20 \,cm$ and $R_{2}=-20 \,cm$
$\therefore \frac{1}{f} =(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) $
$=(1.5-1)\left[\frac{1}{(+20)}-\frac{1}{(-20)}\right]$
$=0.5 \times \frac{2}{20}=\frac{1}{20}$
$ f =20 \,cm$
Incident rays travelling parallel to the axis of lens will converge at its second principal focus. Hence,
$L=+20\, cm$