Answer
(a) increase
(b) $3.5mm $
Work Step by Step
(a) We know that the difference in refraction between the lens-water is smaller than the lens-air system. Thus, when a cormorant plunges into the ocean, its refraction decreases. Thus, the compensating change in the refractive power must increase.
(b) We know that the refractive power before the cormorant enters the ocean is given as
$=\frac{1}{f}$
$=\frac{1}{4.2\times 10^{-3}m}=238.095 diopters $
and the refractive power after the cormorant enters the ocean is given as
$=238.095diopters+45diopters=283.095diopters $
Now the focal length of the cormorant's eye after it enters the ocean is given as
$ f=\frac{1}{283.095diopters}$
$ f=3.5\times 10^{-3}m=3.5mm $
