A) 0.15
B) 0.20
C) 0.25
D) 0.40
Correct Answer: B
Solution :
Given: Focal length of concave lens \[=50\,cm=0.5\,m\] Refractive index of lens of \[{{\mu }_{lens}}=1.5\] Refractive index of liquid \[{{\mu }_{liquid}}=\frac{15}{8}\] Now, from lens markers formula \[\frac{1}{{{f}_{lens}}=}({{\mu }_{lens}}-1)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{1}}} \right)\] \[\Rightarrow \] \[\frac{1}{{{f}_{lens}}}=(1.5-1)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)\] \[\Rightarrow \] \[\frac{1}{{{f}_{lens}}\times 0.5}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}\] \[\Rightarrow \] \[\frac{2}{{{f}_{lens}}}=\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)\] ?(i) Now, the lens is immersed in liquid of refractive index \[\frac{15}{8}\]then, \[\frac{1}{f}=\left( \frac{{{\mu }_{lens}}}{{{\mu }_{liquid}}}-1 \right)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)\] \[\Rightarrow \] \[-\frac{1}{0.5}=\left( \frac{15}{15/8}-1 \right)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)\] \[\Rightarrow \] \[-2=(0.8-1)\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)\] \[\Rightarrow \] \[10=\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)\] ?(ii) Putting the value of \[\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}\]from Eq.(ii) in Eq.(i) , we get \[\frac{2}{{{f}_{lens}}}=10\Rightarrow 10{{f}_{lens}}=2\Rightarrow {{f}_{lens}}=0.2\,m\]
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