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 \[({{\mu }_{l}})\]then its focal length, \[\frac{1}{{{f}_{l}}}=\left( \frac{{{\mu }_{g}}}{{{\mu }_{l}}}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]or \[\frac{1}{{{f}_{1}}}=\frac{({{\mu }_{g}}-{{\mu }_{l}})}{{{\mu }_{l}}}\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] ?(ii) Dividing Eq. (i) by Eq. (ii), we get\[\frac{{{f}_{1}}}{f}=\frac{({{\mu }_{g}}-1){{\mu }_{l}}}{({{\mu }_{g}}-{{\mu }_{l}})}\] ?(iii) But it is given that refractive index of lens is equal to refractive index of liquid ie, \[{{\mu }_{g}}={{\mu }_{l}}.\]Hence, Eq. (iii) gives,\[\frac{{{f}_{l}}}{f}=\frac{({{\mu }_{g}}-1){{\mu }_{l}}}{0}=\infty \](infinity).
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