An object is placed `50 cm` from the surface of a glass sphere of radius `10 cm` along the diameter. Where will the final image be formed after refraction at both the surfaces ? `mu` of glass `=1.5`.

Answer

`20 cm`

Explanation

Here, `u = -50 cm, R = 10 cm, mu_1 = 1, mu_2 = 1.5`
Refraction at surface `P_1 A`
Virtual image at `I_1` where `P_1 I_1 = v_1`
`:. -(mu_1)/(u)+(mu_2)/(v)=(mu_2 - mu_1)/( R)`
`- (1)/(-50)+(1.5)/(v_1) = (1.5 - 1)/(10) = (1)/(20)`
`(3)/(2 v_1)=(1)/(20)-(1)/(50)=(3)/(100), v_1 = 50 cm`
Refraction at surface `P_2 B`
`I_1` acts as virtual object
`u = P_2 I_1 = P_1 I_1 - P_1 P_2 = 50 - 20 = 30 cm`,
`v_2 = P_2 I = ?, R = -10 cm`
`:. -(mu_2)/(u)+(mu_1)/(v)=(mu_1 - mu_2)/( R)`
`-(1.5)/(30)+(1)/(v)=(1 - 1.5)/(-10)=(1)/(20)`
`(1)/(v)=(1)/(20)+(3)/(60)=(1)/(10), v = 10 cm`
Distance of final image `I` from centre of sphere
`CI = CP_2 + P_2 I = 10 + 10 = 20 cm`.
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