Question

# One end of a long glass rod (n 5 1.50) is formed into a convex surface with a radius of curvature of magnitude 6.00 cm. An object is located in air along the axis of the rod. Find the image positions corresponding to object distances of (a) 20.0 cm, (b) 10.0 cm, and(c) 3.00 cm from the convex end of the rod.

Solution
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#### the image within the rod$$\dfrac {n_1}{p} + \dfrac {n_2}{q} = \dfrac {n_2 - n_1}{R} = \dfrac {1}{12}$$(a) $$\dfrac {1}{20} + \dfrac {1.5}{q} = \dfrac {1}{12}$$$$\rightarrow q = 45 cm$$(b) $$\dfrac {1}{10} + \dfrac {1.5}{q} = \dfrac {1}{12}$$$$\rightarrow q = -90 cm$$(c) $$\dfrac {1}{3} + \dfrac {1.5}{q} = \dfrac {1}{12}$$$$\rightarrow q = -6 cm$$

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