Optics

I don't get how this is a mechanics problem.

Anyways, we can imagine a convex mirror in air (where the index of refraction is nearly $1$ ) with rays of light incidence to it that are perpendicular to the plane in which it sits. Here is an illustration

image.

If the air is now replaced with water (which has a greater index of refraction than air), then the light will refract much less. This will increase the focal length.

The 'power' of a lens is the inverse of the focal length. Therefore, if the focal length increases, the power must decrease .

let n1 = refractive index of the surrounding medium n2 = refractive index of the lens according to LENS' LAW 1/f=[(n2/n1) - 1] * k ( where k=radii of each curvature) when the lens is placed in water (n1) will increase because refractive index of water is more than air. therefore, (n2/n1) will decrease as a result [(n2/n1) - 1] will decrease thus (1/f) will also decrease if (1/f) decreases then (f) will obviously increase