A) oil
B) water
C) both (a) and (b)
D) none of the above
Correct Answer: C
Solution :
Key Idea: Refractive index of oil and water both are greater than the refractive index of lens in air. From lens maker's formula \[\frac{1}{f}=(n-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] where\[f\] is focal length of the lens, in the refractive index, \[{{R}_{1}}\]and \[{{R}_{2}}\]the radii of curvature of the surfaces of lens. Also we know that refractive index of oil and water is greater than that for lens. Hence, \[_{l}{{n}_{g}}=\frac{_{a}{{n}_{g}}}{_{a}{{n}_{l}}}<1\] \[\therefore \] \[\frac{1}{{{f}_{1}}}={{(}_{l}}{{n}_{g}}-1)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] that is, focal length of the lens in liquid becomes negative. Hence, the lens which was convex (focal length positive) in air will behave as a concave lens in that liquid. Hence, convex lens becomes less converging in both oil and water.
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