If a thin lens with an index of 1.5 has a dioptric power of +4.00 in air, what is its power if placed in water?

To determine the power of a lens when placed in a medium different from air, we must first understand how the lens formula intersects with the properties of the surrounding medium. The lens formula in this context can be modified to account for the refractive indices of both the lens material and the surrounding medium.

The dioptric power of a lens (P) is given by the formula:

[ P = (n_{lens} - n_{medium}) / f ]

where ( n_{lens} ) is the refractive index of the lens material, ( n_{medium} ) is the refractive index of the surrounding medium (water in this case), and ( f ) is the focal length of the lens.

Given that the index of the lens is 1.5 and the power in air is +4.00 diopters, we can derive the focal length (f) using the relationship:

[ P = \frac{1}{f} ]

Thus,

[ f = \frac{1}{P} = \frac{1}{4} = 0.25 \text{ meters (or 25 cm)} ]

Next, considering water, which has a refractive index of about 1.33