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The radius of curvatureof eachsurface of a convex lens of refractive index 1.5 is 40 cm. Calculate its power. |
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Answer» Solution : Here, `N = 1.5 , R_(1) = +40 cm = 0.04 m` `R_(2) = - 40 cm = - 0.04 m` POWER (p) = `(1)/(F) = (n-1) ((1)/(R_(1)) - (1)/(R_(1)))` `=(1.5-1)[(1)/(0.04)-(1)/((-0.04))]=0.5xx(2)/(0.04)` P = 2.5 D |
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