A) increases by a factor of 1.25
B) increases by a factor of 2.5
C) increases by a factor of 1.2
D) decreases by a factor of 1.2
Correct Answer: B
Solution :
Lens-makers formula is given by \[\frac{1}{f}={{(}_{a}}{{\mu }_{g}}-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] ?.(i) where \[_{a}{{\mu }_{g}}\] 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 \[{{\mu }_{l}}\] then \[\frac{1}{{{f}_{l}}}={{(}_{l}}{{\mu }_{g}}-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] ?..(ii) Here, \[_{l}{{\mu }_{g}}\] is refractive index of glass w.r.t. liquid. Dividing Eq. (i) by Eq. (ii), we have \[\frac{{{f}_{l}}}{f}=\frac{{{(}_{a}}{{\mu }_{g}}-1)}{{{(}_{l}}{{\mu }_{g}}-1)}\] \[\Rightarrow \] \[\frac{{{f}_{l}}}{f}=\left( \frac{1.5-1}{\frac{1.5}{1.25}-1} \right)\] \[\Rightarrow \] \[\frac{{{f}_{l}}}{f}=\frac{0.5\times 1.25}{0.25}=2.5\] Hence, focal length increases by a factor of 2.5.
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