A) Equal to 40 cm
B) Equal to 20 cm
C) Equal to 10 cm
D) None of the above
Correct Answer: A
Solution :
Key Idea: Use Len?s maker?s formula for focal length of lens in different media. If the refractive index of the material of a lens is \[\mu ,\]and \[{{R}_{1}}\] and \[{{R}_{2}}\] are radii of curvatures, then from lens maker?s formula, we have \[\frac{1}{{{f}_{a}}}=\left( _{a}{{\mu }_{g}}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] [in air] \[\frac{1}{{{f}_{\omega }}}=\left( _{\omega }{{\mu }_{g}}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] [in water] \[\Rightarrow \] \[\frac{{{f}_{\omega }}}{fa}=\frac{\left( _{a}{{\mu }_{g}}-1 \right)}{\left( _{\omega }{{\mu }_{g}}-1 \right)}\] \[\Rightarrow \] \[{{f}_{\omega }}=fa\frac{\left( _{a}{{\mu }_{g}}-1 \right)}{\left( _{\omega }{{\mu }_{g}}-1 \right)}={{f}_{a}}\frac{\left( \frac{3}{2}-1 \right)}{\left( \frac{9}{8}-1 \right)}\] \[\Rightarrow \] \[{{f}_{\omega }}=10\times \frac{1}{2}\times \frac{8}{1}=+40\,\,cm\] Note: Since focal length is positive, the nature of convex lens does not change in water \[i.e.\], it is again converging.
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