1.

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.

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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