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The radius of curvature of eachface of biconcave lens, made of glass of refractive index 1.5 is 30 cm. Calculate the focal length of the lens in air. |
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Answer» Solution :Here `n = 1.5 , R_(1) = - 30 CM , R_(2) = 3 cm` Using len.s maker.s formula, `(1)/(F) = (n-1)((1)/(R_(1))-(1)/(R_(2)))` `=(1.5-1)[(1)/(-30)-(1)/(30)]=0.5xx((-2)/(30))` `(1)/(f) = - (1)/(30)` f = - 30 cm |
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