Double-convex lenses are to be manufactured f | vedclass

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(D) Refractive index of glass,$\mu = 1.55$.Focal length of the double-convex lens,$f = 20\; cm$.For a double-convex lens with equal radii of curvature

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(D) Refractive index of glass,$\mu = 1.55$.Focal length of the double-convex lens,$f = 20\; cm$.For a double-convex lens with equal radii of curvature,let $R_1 = R$ and $R_2 = -R$.Using the Lens Maker's formula:$\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} ight]$$\frac{1}{20} = (1.55 - 1) \left[ \frac{1}{R} - \frac{1}{-R} ight]$$\frac{1}{20} = 0.55 	imes \left[ \frac{2}{R} ight]$$\frac{1}{20} = \frac{1.1}{R}$$R = 1.1 	imes 20 = 22\; cm$.Thus,the required radius of curvature is $22\; cm$.

Double-convex lenses are to be manufactured from a glass of refractive index $1.55$,with both faces of the same radius of curvature. What is the radius of curvature (in $cm$) required if the focal length is to be $20\;cm$?

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