InterviewSolution
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InterviewSolution
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A small point objects is placed in air at a distance of `60 cm` from a convex spherical refractive surface of `mu = 1.5`. If radius of curvature of spherical surface is `25 m`, calculate the position of the image and the power of the refracting surface. |
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Answer» Here, `u = -60 cm, mu_(1)= 1, mu_(2) = 1.5, R= + 25 cm, v = ?` As refraction occurs from rarer to denser medium, therefore `(-mu_1)/(u) + (mu_2)/(v) = (mu_2 - mu_1)/( R)` `(-1)/(-60)+(1.5)/(v) = (1.5 - 1)/(25)` `(3)/(2 v) = (1)/(50) - (1)/(60) = (1)/(300)` `v = (300 xx 3)/(2) = 450 cm` As `v` is positive, image formed on the other side of the object, i.e., in refracting denser medium at `45 cm` from the pole. Power of the refracting surface, `P = (mu_2 - mu_1)/( R) = (1.5 - 1)/(0.25) = (0.5)/(0.25) = 2 D`. |
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