The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of the cornea (7.8 mm). This surface separates two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus.

We model the eye as a single spherical refracting surface that separates air from the eye’s interior. The refractive indices are $$n_1 = 1$$ (air) and $$n_2 = 1.34$$ (eye). The radius of curvature of the cornea is given as $$R = 7.8\ \text{mm}$$.

First we convert this radius into centimetres so that every length unit is consistent with the options:

$$R = 7.8\ \text{mm} = 0.78\ \text{cm}.$$

For refraction at a spherical surface we use the standard formula

$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R},$$

where

$$u$$ is the object distance (measured from the surface),
$$v$$ is the image distance that we have to find, and
$$R$$ is positive because the centre of curvature lies in the second medium.

A parallel beam of light corresponds to an object at infinity, so $$u \to \infty$$. Hence

$$\frac{n_1}{u} = \frac{n_1}{\infty} = 0.$$

Substituting this into the formula we get

$$\frac{n_2}{v} = \frac{n_2 - n_1}{R}.$$

Now we solve algebraically for $$v$$:

$$\frac{1}{v} = \frac{n_2 - n_1}{n_2 R} \quad\Longrightarrow\quad v = \frac{n_2 R}{n_2 - n_1}.$$

Substituting the numerical values step by step:

$$v = \frac{1.34 \times 0.78\ \text{cm}}{1.34 - 1}.$$

Compute the denominator first:

$$n_2 - n_1 = 1.34 - 1 = 0.34.$$

Next calculate the product in the numerator:

$$1.34 \times 0.78 = 1.0452.$$

So we have

$$v = \frac{1.0452\ \text{cm}}{0.34}.$$

Carrying out the division:

$$v = 3.0741176\ \text{cm} \approx 3.07\ \text{cm}.$$

Thus the parallel beam is focussed $$3.07\ \text{cm}$$ behind the refracting surface.

Hence, the correct answer is Option D.