A point object is placed at a distance R from the convex surface AB of a glass of radius of curvature `R_(1)=R` and refractive index `n=1.5`. The second curved surface CD is silvered and its radius of curvature is `R_(2)=R` if the distance of the final image formed is `alphaR` from the surface AB rhen find `alpha`?

Answer

2

Explanation

Using `(mu_(2))/(V)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R)` at AB
`(1.5)/(V)+(1)/(2R)=(0.5)/(R)`
`V=infty`
Using `(1)/(V)-(1)/(u)=(1)/(f)` at CD
`(1)/(V)-(1)/(infty)=(1)/((+R//2))`
Again refraction from AB
`(1)/(V)-(1.5)/((-(3R)/(2)))=(1-1.5)/((-R))`
`(1)/(V)+(1)/(R)=(1)/(2R)`
`(1)/(V)=-(1)/(2R)`
`V=-2R`