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\[({{\mu }_{t}})\]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 \[\frac{1}{{{f}_{l}}}=\frac{({{\mu }_{g}}-{{\mu }_{l}})}{{{\mu }_{l}}}\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] ?. (ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{{{f}_{l}}}{f}=\frac{({{\mu }_{g}}-1){{\mu }_{l}}}{({{\mu }_{g}}-{{\mu }_{l}})}\] ? (iii) But it is given that refractive index of lens is equal to refractive index of liquid 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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