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A biconcave lens made of transparent material of refractive index `1.25` is immersed in water of refractive index `1.33`. Will the lens behave as a convering or diverging lens? Given reason. |
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Answer» `(1)/(f)=((mu_(i))/(mu_(s))-1) [(1)/(R_(1))-(1)/(R_(2))]` For concave lens `[(1)/(R_(1))-(1)/(R_(2))]=-ve` here `=1.25` `s=1.32` Then `((mu_(i))/(mu_(s))-1)=-ve` Then `(1)/(f)rarr(-ve)(-ve)` Thus it will behave like converging lens. |
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