In This post you will get answer of Q.
A lens behaves as a converging lens in air and diverging lens in water. The refractive index of the material of the lens is
Solution:
If a lens (made of glass) of refractive index
$ {{mu }_{g}} $ is immersed in a liquid of refractive index
$ {{mu }_{l}} $
then its focal length in liquid
$ {{f}_{l}} $ is given by
$ frac{1}{{{f}_{l}}}={{(}_{l}}{{mu }_{g}}-1)left[ frac{1}{{{R}_{1}}}-frac{1}{{{R}_{2}}} right] $ ..(i)
If $ {{mu }_{g}}
then $ {{f}_{l}} $ and $ {{f}_{a}} $
have opposite signs and the nature of lens changes ie, convex lens diverges the light rays and concave lens converges the light rays.
Solution:
If a lens (made of glass) of refractive index
$ {{mu }_{g}} $ is immersed in a liquid of refractive index
$ {{mu }_{l}} $
then its focal length in liquid
$ {{f}_{l}} $ is given by
$ frac{1}{{{f}_{l}}}={{(}_{l}}{{mu }_{g}}-1)left[ frac{1}{{{R}_{1}}}-frac{1}{{{R}_{2}}} right] $ ..(i)
If $ {{mu }_{g}}
then $ {{f}_{l}} $ and $ {{f}_{a}} $
have opposite signs and the nature of lens changes ie, convex lens diverges the light rays and concave lens converges the light rays.