Do Things Seem 33% Bigger Underwater?

Not quite. Refraction at a planar water-air interface (such as due to a mask faceplate) produces an image that seems larger than the corresponding object, but the degree of magnification depends on both the object-to-faceplate and eye-to-faceplate distances. If the index of refraction of the water is n(w)=1.33 and that of air is n(a)=1.00, the factor 1.33 is the maximum possible magnification that can be produced by a planar water-air interface.

In the diagram shown below, a simplified version of the eye (the reduced eye) has been represented as consisting of an aqueous humour, and a single spherical refracting surface with radius of curvature R, which focusses the light rays from an object of height 'h' onto an image on the retina a distance 'I' from the surface. The figure shows the usual result of ray optics in air.

By similar triangles, the height of the retinal image is

             h(a) = h(I-R)/(O+R)                             Equation 1

For the same relative positions of object and eye, but with the object in water and the eye in air behind a water-air interface, the configuration is as shown below. Note the refraction at the faceplate, with angle of incidence 'i' and angle of refraction 'r', and the magnified retinal image.

The image height is now

             h(w)=(I-R)tan(r)                                Equation 2

Again using similar triangles, the object height can be expressed as

             h = Dtan(i) + (d+R)tan(r)                       Equation 3

Substituting this expression for h into Equation 1, and then using Equation 2, we get the ratio

             h(w)/h(a) = (O+R)/[Dtan(i)/tan(r) + (d+R)]      Equation 4

But the refraction at the faceplate is described mathematically by Snell's Law, i.e.

             sin(i)/sin(r) = n(a)/n(w)                       Equation 5

and for small angles at the eye (paraxial rays) we know that the sines and tangents of angles are approximately the same, so we can assume that

             tan(i)/tan(r) = n(a)/n(w)                       Equation 6

Substituting this into Equation 4 and noting that O=D+d, we get the retinal image magnification

             M = h(w)/h(a) = (D+d+R)/[Dn(a)/n(w) + (d+R)]    Equation 7

or, in the more instructive form,

             M =m{[ (1+ (d+R)/D]/[1+m(d+R)/D]}               Equation 8


             m = n(w)/n(a) = 1.33                            Equation 9

It is now clear that the magnification M due to the water-air interface is always LESS than 1.33. M approaches this value only in the limit as the distance D (object to mask) approaches infinity, or when both d and R approach zero. The following table gives some representative values of M for typical values of D and d, with R=0.8 cm

      object to   eye to    
      faceplate  faceplate
       D (cm)     d (cm)       M
        200         0        1.328
        100         0        1.327
         50         0        1.323
        200        2.5       1.323
        100        2.5       1.316
         50        2.5       1.303
         25        2.5       1.281
        100        5.0       1.306
         50        5.0       1.286
         25        5.0       1.252

As the object approaches the faceplate and/or the eye recedes from it, the magnification decreases. At an object distance of 25 cm with the eye 2.5 cm behind the faceplate, the image is only 28% larger.

The foregoing dealt strictly with the size of the retinal image. How the brain perceives that image depends on many other factors. Studies have shown (Ross et al), that the brain adapts rather quickly to the underwater environment, so that initial errors in judging object sizes are significantly reduced during the course of a dive. Moreover, size judgments improve with diving experience.


H. E. Ross, S. S. Franklin, G. Weltman, and P. Lennie, 'Adaptation of Divers to Size Distortion Under Water', Br. J. Psychol. (1970), 61, pp. 365-373.