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.

Explanation

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 occurs 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)/(2 v) = (1)/(6)-(1)/(12)=(1)/(12)`
`v = 18 cm`
As `v` is positive, the image is real.
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