A)
B)
C)
D)
Correct Answer: A
Solution :
If a lens of refractive index m is immersed in a medium of refractive index m1, then its focal length in medium is given by \[\frac{1}{{{f}_{m}}}=\left( _{m}{{\mu }_{l}}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\] If \[{{f}_{a}}\] is the focal length of lens in air, then \[\frac{1}{{{f}_{a}}}=\left( _{a}{{\mu }_{l}}-1 \right)\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]\[\Rightarrow \]\[\frac{{{f}_{m}}}{{{f}_{a}}}=\frac{\left( _{a}{{\mu }_{l}}-1 \right)}{\left( _{m}{{\mu }_{l}}-1 \right)}\] If \[{{\mu }_{1}}>\mu ,\]then\[{{f}_{m}}\] and \[{{f}_{a}}\] have opposite signs and the nature of lens changes i.e. a convex lens diverges the light rays and concave lens converges the light rays. Thus given option (a) is correct.
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