The radius of curvature of the convex face of a plano-convex lens is`12`cm and its refractive index is`1.5`(a)Find the focal length of this lens.The plane surface of the lens is now silvered.(b)AT what distanc efrom the lens will parallel rays incident on the convex face converage?(c)Sketch the ray diagram to locate the image,when a point object is placed on the axis`20cm`from the lens(d)Calculate the image distance when the object is placed as in (c).

Explanation

(a)As for a lens,by lens -maker's formula
`(1)/(f)=(mu-1)[(1)/(R_(1))-(1)/(R_(2))]`
Here `mu=1.5,R_(1)=12cm`and `R_(2)=oo`
So`(1)/(f)=(1.5-1)[(1)/(12)-(1)/(oo)]`i.e,`f=24cm`
i.e,the lens as convergent with focal length`24cm`

(b)As light after passing through the lens will be incident on the mirror which will reflect it back throught the lens again,so
`P=P_(L)+P_(M)+P_(L)=2P_(L)+P_(M)`
`P_(L)=(1)/(f_(L))=(1)/(0.24)` and `P_(M)=-(1)/(oo)=0` [as `f_(M)(R)/(2)=oo`]
`P=2xx(1)/(0.24)+0=(1)/(0.12)D`
The system is equivalent to a concave mirror of focal length`F_(1)`
`p=-(1)/(F) i.e,F=-(1)/(P)=-0.12m=-12cm`
i.e,the rays will behave as a concave mirror of focal length `12`cm.So as for parallel incident rays`u=-oo`from mirror formula`(1)/(v)+(1)/(u)=(1)/(f)` we have `(1)/(v)+(1)/(oo)=-(1)/(12)`i.e,`v=-12`
i.e, parallel incident rays will focus will at a distance of `12cm`in front of the lens as shown in fig.(c)and(d) when object is at`20cm` in front of the given silvered lens which behaves as a concave mirror of focal length`12cm`from mirror formula
`(1)/(v)+(1)/(u)=(1)/(f)` we have
`(1)/(v)+(1)/(-20)=(1)/(-12)`i.e,`v=-30`cm
i.e, the silvered lens will form image at a distance of `30`cm in front of it as shown in fig.(c ).