A point object O is placed on the axis of a cylindrical piece of glass of refractive index 1.6 as shown in the figure. One surface of the glass piece is convex with radius of curvature 3mm. The point appeared to be at 5 mm on the axis when viewed along the axis and from right side of convex surface. The distance of the point object from the convex surface is :

Explanation

The head appears to be `5.0mm` into the amber only because the light rays that the observer intercepts are bent by refraction at the convex amber surface. The image distance `i` differes from the object distance p according to Eq. 34-11 . To use that euation to find the object distance , we first note :
1. Because the object (the head) and its image are on the same side of the refracting surface, the image must be virtual and so `i=-5.0mm`
2. Because the object is always taken to be in the medium of index of refraction `n_(1)`, we must have `n_(1)=1.6` and `n_(2)=1.0`
3. Because the object faces a concave refracting surface, the radius of curvature r is negative and so `r=-3.0mm`.
Calculations : Making these substitutions in Eq. 34-11
`(n_(1))/(p)+(n_(2))/(i)=(n_(2)-n_(1))/(r )`,
yields `(1.6)/(p)+(1.0)/(-5.0mm)=(1.0-1.6)/(-3.0mm)`
and `p=4.0mm`