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

(b) As light after passing through the lens will be incident on the mirror which will reflect it back through 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 ["as" f_(M) = (R)/(2) = oo]`
So `P = 2 xx(1)/(0.24) + 0 = (1)/(0.12) D`.
The system is squivalent to a concave mirror of focal length `F , P = -(1)/(F)`
i.e., `F = -(1)/(P) = -0.12"m" = -12"cm" ` 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)`
`implies v=-12` cm i.e., parallel incident rays will focus will at a distance of `12` cm in front of the lens as shown in Figure (c) and (d) 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)/(f)`
we have `(1)/(v) + (1)/(-20) = (1)/(-12)`
`implies 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)`