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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?

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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.

Posted by

Pankaj Sanodiya

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