A point source is located at a distance of `20 cm` from the front surface of a symmetrical glass biconvex lens. The lens is `5.0 cm` thick and the curvature radius of its surfaces is `5.0 cm`. How far beyond the rear surface of this lens is the image of the source fromed ?

Explanation

As the given lense has significant thickness, the lense, formula cannot be used.
For refraction at the front surface from the formula `(n')/(s') - (n)/(s) = (n' - n)/(R )`
`(1.5)/(s') - (1)/(-20) = (1.5 - 1)/(5)`
On simplifying we get, `s' = 30 cm`.
Thus the image `I`' produced by the front surface behaves as a virtual source for the rear surface at distance `25 cm` from it, because the thickness of the lense is `5 cm`. again from the refraction fromula at cerve surface
`(n')/(s') - (n)/(s) = (n' - n)/(R )`
`(1)/(s') - (1.5)/(25) = (1- 1.5)/(-5)`
On simplifying, `s' =+6.25 cm`
Thus we get a real image `I `at a distance `6.25cm` beyond the rear surface (Fig.).