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Question
In the given figure the radius of curvature of the curved face in the planoconvex and the planoconcave lens is 15 cm each. The refractive index of the material of the lenses is 1.5. Find the final position of the image formed.
Solution
The focal length of the plano-convex lens is:
`1/f = (1.5 - 1)(1/((+10)) - 1/∞) = 1/20`
`f = 20`
The focal length of plano-concave lens is:
`1/f = (1.5 - 1)(1/∞ - 1/10) = -1/20`
`f = -20`
Since parallel beams are incident on the lens, its image from plano-concave lens will be formed at +30 cm from it (at the focus) and will act as an object for plano-concave lens. Since the two lenses are separated by a distance of 20 cm, therefore for the next lens.
u = +10 cm
`v = (uf)/((u + f)) = (10 xx(- 20))/(10 + (- 20)`
`v = (-200)/(-10)`
`v` = 20 cm (final position of the image formed)
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