The convex surface of a plano-convex glass lens with curvature radius `R = 40 cm` comes into contact with a glass plate. A certain ring observed in reflected light has a radius `r = 2.5 mm`. Watching the given ring, the lens was gradually removed from the plate by a distance `Delta h = 5.0 mu m`. What has the radius of that ring become equal to?

Explanation

The path traversed in air film of the wave constituting the `K^(th)` ring is
`(r^(2))/(R ) = (1)/(2) k lambda`
when the lens is moved a distance `Deltah` the ring radius changes to `r`' and the path length becomes
`(r'^(2))/(R ) + 2Deltah = (1)/(2)k lambda`
Thus `r' = sqrt(r^(2) - 2R Delta h) = 1.5 mm`.