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A lens is made of flint glass (refractive index `=1.5`). When the lens is immersed in a liquid of refractive index `1.25` , the focal length:A. increases by a factor of `1.25`B. increases by a factor of `2.5`C. increase by a factor of `1.2`D. decrease by a factor of `1.2` |
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Answer» Correct Answer - b Lens-maker formula given by `(1)/(f)= (._(a)mu_(g)-1) ((1)/(R_(1))-(1)/(R_(2)))`….(i) where `_amu_(g)` is refractive index of glass `w.r.t` air, `R_(1)` and `R_(2)` are radii of curvature of two surfaces of lens and `f` is focal length of the lens. If the lens is immersed in a liquid of refractive index `mu_(1)` then `(1)/(f_(l))=(._(l)mu_(g)-1) ((1)/(R_(1))-(1)/(R_(2)))`....(ii) Here, `_(l)u_(g)` is refrective index of glass w.r.t liquid. Dividing Eq.(i) by Eq.(ii) we have `(f_(1))/(f)=((._(a)mu_(g)-1))/((._(l)mu_(g)-1))implies (f_(l))/(f)((1.5-1)/((1.5)/(1.25)-1))` `implies (f_(l))/(f)= (0.5xx1.25)/(0.25)= 2.5` Hence, focal length increases by a factor of `2.5`. |
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