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A double convex lens made of glass of refractive index 1.5 has both radii of curvature 20 cm each. Find the focal length of the lens. If an object is placed at a distance of 15 cm from this lens, find the position of the image formed. |
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Answer» Solution :Here `n = 1.5 , R_(1) = + 20 cm , R_(2) = - 20cm` `(1)/(f) = (n-1) ((1)/(R_(1)) - (1)/(R_(2)))` ` = (1.5 - 1) [(1)/(20) - (1)/(-20)] = 0.5 XX ((2)/(20))` `(1)/(f) = (1)/(20)` f = 20 cm Now, u = - 10 cm and f = + 20 cm From THIN lens formula, `(1)/(v) = (1)/(f) + (1)/(u) = (1)/(20) - (1)/(10) = - (1)/(20)` `therefore` v = - 20 cm MAGNIFICATION, m = `(h_(2))/(h_(1)) = (v)/(u)` `(h_(2))/(5cm) = (-20)/(-10)` Hence a VIRTUAL and erect image of HEIGHT 10 cm is formed at a distance of 20 cm from the lens on the same side as the object. |
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