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Find the power of a thin glass lens (mu = 1.5) in a liquid with refractive index mu_(0) = 1.7, if its power in air is : -5D. |
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Answer» SOLUTION :`(b) P_(0)=(mu-1)[(1)/(R_(1))-(1)/(R_(2))]` in air `P=(mu-mu_(0))[(1)/(R_(1))-(1)/(R_(1))]"in liquid"` `therefore P=P_(0)(mu-mu_(0))/(mu-1)=-5xx(1.5-1.7)/(1.5-1)=+2D` |
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