A) equal to unity
B) below 1.33
C) greater than 1.33
D) less than unity
Correct Answer: B
Solution :
Key Idea: Focal length of diverging lens is negative. The focal length\[f\]of a lens depends upon the refractive index n of the material of lens and radii of curvatures of the lens as \[\frac{1}{{{f}_{l}}}={{(}_{l}}{{n}_{g}}-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] When refractive index of the liquid is greater than the refractive index of the material of the lens then \[{{(}_{a}}{{n}_{l}}{{>}_{a}}{{n}_{g}})\]then \[_{l}{{n}_{g}}=\frac{_{a}{{n}_{g}}}{_{a}{{n}_{g}}}<1\] \[\frac{1}{{{f}_{1}}}={{(}_{l}}{{n}_{g}}-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] That is, the focal length of the lens becomes negative. Hence, the lens which was convex (focal length positive) in air will behave as a concave lens in liquid\[(\mu =1.33)\]. Hence, refractive index of lens is below 1.33
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