An object is placed at a distance of 10 cm in front of a double convex lens made of glass of refractive index 1.5. Both the radii of curvature of the lens are 30 cm in magnitude. What is the position of the image formed?
Given, Distance of the object from the lens = u = -10 cm Refractive index of the lens = µ = 1.5 Radii of curvature of the lens are 30 cm in magnitude R1 = 30 cm and R2 = -30 cm (As per sign convention) According to Len's Maker's formula 1/f = (µ - 1)(1/R1 - 1/R2) = (1.5 - 1)(1/30 - (-1/30) = 0.5 x 2/30 = 1/30 or, f = 30 cm From lens equation, 1/v - 1/u = 1/f v = fu/(u + f) = (30 x -10)/(-10 + 30) = -300/20 = -15 cm The image is formed 15 cm on the same side as the object
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