Question

A glass dumbbell of length `30 cm` and refractive index `1.5` has ends of radius of curvature `3cm`. A point object is situated at a distance of `12 cm` from one end of dumbbell. Find the position of the image formed due to refraction ai one end only.

Answer 1

Here, `l = 30 cm, mu_(2) = 1.5, mu_(1) = 1`,
`R = 3 cm, u = -12 cm`.
Image formed is at `I` , as shown in Fig.
`PI = v = ?`

As refraction sccurs from air to glass.
`:. -(mu_(1))/(u) + (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )`
`-(1)/(-12) + (1.5)/(v) = (1.5 - 1)/(3) = (1)/(6)`
`(3)/(2v) = (1)/(6) - (1)/(12) = (1)/(12)`
`v = 18 cm`
As `v` is positive, the image is real.