The redius of curvature of the convex face of a plano-convex lens is 12cm 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 distance form the lans will parallel rays incident on the convex face converge?
c. Sketch the ray diagram to locate the image, when a point object is places on the axis 20cm from the lens.
d. Calculate the image distance when the object is placed as in (c).

Explanation

a. As far a lens, by a lensmarker's formula,
`(1)/(f)=(mu_1)[(1)/(R_(1))-(1)/(R_(1))]`
Here `mu=(1.5-1),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 is 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)`
But `P_(L)=(1)/(f_(L))=(1)/(0.24) and P_(M)=-(1)/(oo)=0` `[f_(M)=(R)/(2)=oo]`
So, `P=2xx(1)/(0.24)+0=(1)/(0.12)D`
The system is equibalent to a concave mirror of focal length F.
`P=-(1)/(F)`i.e.,
`F=-(1)/(P)=-0.12m=-12cm`
i.e., the system will behave as a concave mirror of focal length 12cm . 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)rArrv=-12cm`
i.e., parallel incident rays will focus at a distance of 12cm in front of the lens as shown in Figure. When object is at 20 cm in front of the given silvered lens, which behaves as a concave mirror of focal length
12 cm, from mirror formula `(1)/(v)+(1)/(u)=(1)/(12)rArrv=-12cm`
`(1)/(v)+(1)/(-20)=(1)/(-12)rArrv=-30cm `
i.e., the silvered lens will form an image at a distance of 30cm in front of it as shown in Figure.