A) \[\frac{17}{8}\]
B) \[\frac{15}{8}\]
C) \[\frac{13}{8}\]
D) \[\frac{9}{8}\]
Correct Answer: B
Solution :
The focal length of a convex lens of refractive index \[{{\mu }_{g}}\] in air is \[\frac{1}{{{f}_{air}}}=({{\mu }_{g}}-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] ?(i) Where \[{{R}_{1}}\]and \[{{R}_{2}}\]are the radius of curvatures of its first and second surface. When lens immersed in a liquid of refractive index\[{{\mu }_{l}}\]then refractive index of material of lens (glass) with respect to liquid is \[_{l}{{\mu }_{g}}=\frac{{{\mu }_{g}}}{{{\mu }_{l}}}\] ?(ii) \[\therefore \] Focal length of lens in liquid is \[\frac{1}{f}={{(}_{l}}{{\mu }_{g}}-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] ...(iii) Dividing (i) by (ii), we get \[\frac{f}{{{f}_{air}}}=\frac{{{\mu }_{g}}-1}{{{(}_{l}}{{\mu }_{g}}-1)}=\frac{{{\mu }_{g}}-1}{\left( \frac{{{\mu }_{g}}}{{{\mu }_{l}}}-1 \right)}\] Putting \[f=-0.5\,m\] \[{{f}_{air}}=0.2\,m\] \[_{a}{{\mu }_{g}}-1=1.5\] \[{{\mu }_{l}}=?\] \[=\frac{-0.5}{0.2}=\frac{1.5-1}{\left( \frac{1.5}{{{\mu }^{l}}}-1 \right)}=\frac{0.5}{\left( \frac{1.5}{{{\mu }^{l}}}-1 \right)}\] \[\Rightarrow \] \[\frac{1.5}{{{\mu }_{l}}}-1=-0.2\] \[\Rightarrow \] \[\frac{1.5}{{{\mu }_{l}}}1-0.2=0.8\] \[{{\mu }_{l}}=\frac{1.5}{0.8}\] \[{{\mu }_{l}}=\frac{15}{8}\] \[\ \therefore \]Refractive index of liquid \[=\frac{15}{8}\]
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