A double convex thin lens made of glass (refractive index $\mu\ = \ 1.5$) 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

Question

Dr. Abubacker · Accepted Answer

(c) The parallel rays coverage at a distance equal to focal length of lens so $f = L$
$but\ \frac{1}{f} = (\mu - 1)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} ight)$
$= (1.5 - 1)\left( \frac{1}{20} + \frac{1}{20} ight)$
$= \frac{0.5 	imes 2}{20} = \frac{1}{20}$
So $f = 20\ cm\ $therefore, $L = 20\ cm$

Q. A double convex thin lens made of glass (refractive index $\mu\ = \ 1.5$) 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

Explanation

Reviewed by Dr. Abubacker

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