Question

Two concave refracting surface of equal radii of curvature and refractive index 1.5 face each other in air as shown in figure. A point is placed midway in between the centre and one of the vertices. What is the separation between the image of O formed by each refracting surface ?

Answer 1

We have for the image formed by the surface on the right side.
`mu_(1) = 1, mu_(2) = 1.5, u = - R//2, R_(1) = -R`
`(mu_(2))/(v) - (mu_(1))/(u) = (mu_(2)-mu_(1))/(2)`
`(1.5)/(v) - (2)/(-R) = (1.50-1.00)/(-R)`
`:. v = -(3R)/(5)`
The image is at a distance of (R - 3R/5) = 0.4 R from the centre P towards the right side. For the surface on the left side,
`mu_(1) = 1, mu_(2) = 1.5, u = 3R//2, R_(2) = R`
`(mu_(2))/(v) - (mu_(1))/(u) = (mu_(2) - mu_(1))/(R_(2))`
`(1.5)/(v) - (2)/(-3R) = (1.50 - 1.00)/(R)`
`(1.5)/(v) = (1)/(2R) + (2)/(3R) = (7)/(6R)`
`:. v = (9R)/(7)`
Hence, the image is at distance of `((9R)/(7-R) = (2R)/(7))` from the centre O towards the right hand side. The distance between the two images = 0.4R - 2R/7 = 0.114R.