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

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

Ray Optics and Optical Instruments

Solution:

Lens maker s formula, $\frac{1}{f} = \left(\mu-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) $
Here, $f = 20 \,cm, \mu= 1.55$,
$ R_{1} =R$ and $R_{2} = -R $
$ \therefore \frac{1}{20}=\left(1.55-1\right)\left(\frac{1}{R }-\frac{1}{\left(-R\right)}\right) $or
$\frac{1}{20}=0.55\times\frac{2}{R}$
$ \Rightarrow R = 1.1 \times 20 $
$= 22 \,cm$