# Q 9.7  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 required if the focal length is to be 20cm?

As we know the lens makers formula

$\frac{1}{f}=(\mu _{21}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

[ This is derived by considering the case when the object is at infinity and image is at the focus]

Where $f$ = focal length of the lens

$\mu _{21}$ = refractive index of the glass of lens with the medium(here air)

$R_1$ and $R_2$ are the  Radius of curvature of faces of the lens.

Here,

Given, $f$ = 20cm,

$R_1$  = $R$ and $R_2$  =  $-R$

$\mu _{21}$ = 1.55

Putting these values in te equation,

$\frac{1}{20}=(1.55-1)(\frac{1}{R}-\frac{1}{-R})$

$\frac{2}{R}=\frac{1}{20}*\frac{1}{0.55}$

$R = 40*0.55$

$R = 22cm$

Hence Radius of curvature of the lens will be 22 cm.

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