Explanation
Power of a lens $$=\dfrac{1}{f}$$
Where $$f $$ is the focal length of the lens.
So, by lens maker formula
$$\dfrac{1}{f}=\left ( \dfrac{\mu _{2}}{\mu _{1}}-1 \right )\left ( \dfrac{1}{R_{1}}-\dfrac{1}{R_{2}} \right )$$
As lens is dipped in water, $$\mu$$ of the medium is increased from 1 to $$\dfrac{4}{3} $$ (i.e. refractive index of water), so, $$\dfrac{\mu_{2}}{\mu_{1}}$$ term decreases, so, $$\left ( \dfrac{\mu_{2}}{\mu_{1}}-1 \right )$$ term decrease which leads to decrease of $$\dfrac{1}{f}.$$ so, power $$\left ( =\dfrac{1}{f} \right )$$ decreases.
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