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Using the following data, calculate refractive index of the material of convex lens. Radii of curvature of both the surfacea are 0.2 m.Data recorded in the displacement//shift method to determine focal length of convex lens is shown here. |
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Answer» Solution :`f_1 = (D_1^2 - x_1^2)/(4 D_1) = (0.82^2 - 0.128^2)/(4 XX 0.82) = 0.2 m` `f_2 = (D_2^2 - x_2^2)/(4 D_2) = (0.9^2 - 0.3^2)/(4 xx 0.9) = 0.2 m` `:.` AVERAGE FOCAL length, `f - (f_1 + f_2)/(2) = (0.2 + 0.2)/(2) = 0.2 m` Now, `R_1 = 0.2 m, R_2 = -0.2 m, mu = ?` From `(1)/(f) = (mu - 1)((1)/(R_1) - (1)/(R_2))` `(1)/(0.2)=(mu - 1) ((1)/(0.2) + (1)/(0.2)) = (mu - 1)/(0.1)` `mu - 1 = (1)/(2), mu = (1)/(2) + 1 = 1.5`. |
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