A) become small, but non-zero
B) remain unchanged
C) become zero
D) become infinite
Correct Answer: D
Solution :
From lens maker?s formula \[\frac{1}{f}=({{\mu }_{g}}-1)\,\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)....(i)\] When convex lens is dipped in a liquid of refractive index \[({{\mu }_{1}})\]then its focal length \[\frac{1}{{{f}_{l}}}=\left( \frac{{{\mu }_{g}}}{{{\mu }_{1}}}-1 \right).\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] \[or\frac{1}{{{f}_{1}}}=\frac{({{\mu }_{g}}-{{\mu }_{l}})}{{{\mu }_{t}}}\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] ....(ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{{{f}_{l}}}{f}=\frac{({{\mu }_{g}}-1){{\mu }_{l}}}{({{\mu }_{g}}-{{\mu }_{l}})}\] ....(iii) But it is given that refractive index of lens is equal to refractive index of liquid i.e., \[{{\mu }_{g}}={{\mu }_{l}}\]. Hence, Eq. (iii) gives, \[\frac{{{f}_{l}}}{f}=\frac{({{\mu }_{g}}-1){{\mu }_{l}}}{0}=\,\infty \] (infinity)
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