A) \[-20\,cm\]
B) \[-25\,cm\]
C) \[-50cm\]
D) \[50cm\]
Correct Answer: C
Solution :
Given \[{{\mu }_{g}}=1.5\] \[{{\mu }_{oil}}\,=1.7\] \[R=20cm\] From Lens Maker's formula for the piano convex lens \[\frac{1}{f}=(\mu -1)\left[ \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right]\] Here, \[{{R}_{1}}=R\] and for plane surface \[{{R}_{2}}=\infty \] \[\therefore \] \[\frac{1}{{{f}_{lens}}}=(1.5-1)\left( \frac{1}{R}-0 \right)\] \[\Rightarrow \] \[\frac{1}{{{f}_{lens}}}\,=\frac{0.5}{R}\] When the intervening medium is filled with oil, then focal length of the concave lens formed by the oil \[\frac{1}{{{f}_{\text{concave}}}}=(17-1)-\left( -\frac{1}{R}-\frac{1}{R} \right)\] \[=-0.7\times \frac{2}{R}=\frac{-14}{R}\] Here, we have two concave surfaces So. \[\frac{1}{{{f}_{eq}}}=2\times \frac{1}{f}+\frac{1}{f}\] \[=2\times \frac{0.5}{R}+\left( \frac{-14}{R} \right)=\frac{1}{R}-\frac{14}{R}=-\frac{0.4}{R}\] \[\therefore \] \[{{f}_{eq}}=-\frac{R}{0.4}=-\frac{20}{0.4}=-50cm\]You need to login to perform this action.
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