9.5 Refraction at Spherical Surfaces and by Lenses

We have so far considered refraction at a plane interface. We shall now consider refraction at a spherical interface between two transparent media. An infinitesimal part of a spherical surface can be regarded as planar and the same laws of refraction can be applied at every point on the surface. Just as for reflection by a spherical mirror, the normal at the point of incidence is perpendicular to the tangent plane to the spherical surface at that point and, therefore, passes through its centre of curvature. We first consider refraction by a single spherical surface and follow it by thin lenses. A thin lens is a transparent optical medium bounded by two surfaces; at least one of which should be spherical. Applying the formula for image formation by a single spherical surface successively at the two surfaces of a lens, we shall obtain the lens maker’s formula and then the lens formula.

9.5.1 Refraction at a spherical surface

Figure 9.17 shows the geometry of formation of image I of an object O on the principal axis of a spherical surface with centre of curvature C, and radius of curvature R. The rays are incident from a medium of refractive index $n_1$ , to another of refractive index $n_2$ . As before, we take the aperture (or the lateral size) of the surface to be small compared to other distances involved, so that small angle approximation can be made. In particular, NM will be taken to be nearly equal to the length of the perpendicular from the point N on the principal axis. We have, for small angles,

$$\begin{array}{l}\tan \mathrm{\angle }\mathrm{NOM}=\frac{\mathrm{MN}}{\mathrm{OM}}\\ \tan \mathrm{\angle }\mathrm{NCM}=\frac{\mathrm{MN}}{\mathrm{MC}}\\ \tan \mathrm{\angle }\mathrm{NIM}=\frac{\mathrm{MN}}{\mathrm{MI}}\end{array}$$

Now, for ∆NOC, i is the exterior angle. Therefore, i = ∠NOM + ∠NCM

$$i=\frac{\mathrm{MN}}{\mathrm{OM}}+\frac{\mathrm{MN}}{\mathrm{MC}}$$ Similarly, $$\begin{array}{c}r=\mathrm{\angle }\mathrm{NCM}-\mathrm{\angle }\mathrm{NIM}\\ \mathrm{i}.\mathrm{e}.,r=\frac{\mathrm{MN}}{\mathrm{MC}}-\frac{\mathrm{MN}}{\mathrm{MI}}\end{array}$$

Now, by Snell’s law $$n_1\sin i=n_2\sin r$$ or for small angles $$n_{1}i=n_{2}r$$

Substituting i and r from Eqs. (9.13) and (9.14), we get

$$\frac{n_1}{OM}+\frac{n_2}{MI}=\frac{n_2-n_1}{MC}$$

Here, OM, MI and MC represent magnitudes of distances. Applying the Cartesian sign convention, $$\mathrm{OM}=-u,\mathrm{MI}=+v,\mathrm{MC}=+R$$ Substituting these in Eq. (9.15), we get

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

Equation (9.16) gives us a relation between object and image distance in terms of refractive index of the medium and the radius of curvature of the curved spherical surface. It holds for any curved spherical surface.

9.5.2 Refraction by a lens

Figure 9.18(a) shows the geometry of image formation by a double convex lens. The image formation can be seen in terms of two steps: (i) The first refracting surface forms the image I 1 of the object O [Fig. 9.18(b)]. The image I 1 acts as a virtual object for the second surface that forms the image at I [Fig. 9.18(c)]. Applying Eq. (9.15) to the first interface ABC, we get

$$\frac{n_1}{\mathrm{OB}}+\frac{n_2}{{\mathrm{BI}}_1}=\frac{n_2-n_1}{{\mathrm{BC}}_1}$$

A similar procedure applied to the second interface* ADC gives,

$$-\frac{n_2}{{\mathrm{DI}}_1}+\frac{n_1}{\mathrm{DI}}=\frac{n_2-n_1}{{\mathrm{DC}}_2}$$

For a thin lens, B$I_1$ = D$I_1$ . Adding Eqs. (9.17) and (9.18), we get

$$\frac{n_1}{\mathrm{OB}}+\frac{n_1}{\mathrm{DI}}=(n_2-n_1)(\frac{1}{{\mathrm{BC}}_1}+\frac{1}{{\mathrm{DC}}_2})$$

Suppose the object is at infinity, i.e., OB → ∞ and DI = f, Eq. (9.19) gives

$$\frac{n_1}{f}=(n_2-n_1)(\frac{1}{B{C}_1}+\frac{1}{D{C}_2})$$

The point where image of an object placed at infinity is formed is called the focus F, of the lens and the distance f gives its focal length. A lens has two foci, F and F′, on either side of it (Fig. 9.19). By the sign convention, $$\begin{array}{l}{\mathrm{BC}}_1=+R_1\\ {\mathrm{DC}}_2=-R_2\end{array}$$ So Eq. (9.20) can be written as

$$\frac{1}{f}=(n_{21}-1)(\frac{1}{R_1}-\frac{1}{R_2})(\because n_{21}=\frac{n_2}{n_1})$$ Lens maker’s formula

It is useful to design lenses of desired focal length using surfaces of suitable radii of curvature. Note that the formula is true for a concave lens also. In that case $R_1$ is negative, $R_2$ positive and therefore, f is negative.

From Eqs. (9.19) and (9.20), we get

$$\frac{n_1}{\mathrm{OB}}+\frac{n_1}{\mathrm{DI}}=\frac{n_1}{f}$$

Again, in the thin lens approximation, B and D are both close to the optical centre of the lens. Applying the sign convention, BO = – u, DI = +v, we get

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$

Equation (9.23) is the familiar thin lens formula. Though we derived it for a real image formed by a convex lens, the formula is valid for both convex as well as concave lenses and for both real and virtual images.

It is worth mentioning that the two foci, F and F′, of a double convex or concave lens are equidistant from the optical centre. The focus on the side of the (original) source of light is called the first focal point, whereas the other is called the second focal point.

To find the image of an object by a lens, we can, in principle, take any two rays emanating from a point on an object; trace their paths using the laws of refraction and find the point where the refracted rays meet (or appear to meet). In practice, however, it is convenient to choose any two of the following rays:

1. A ray emanating from the object parallel to the principal axis of the lens after refraction passes through the second principal focus F′ (in a convex lens) or appears to diverge (in a concave lens) from the first principal focus F.
2. A ray of light, passing through the optical centre of the lens, emerges without any deviation after refraction.
3. (a) A ray of light passing through the first principal focus of a convex lens [Fig. 9.19(a)] emerges parallel to the principal axis after refraction.
   
   (b) A ray of light incident on a concave lens appearing to meet the principal axis at second focus point emerges parallel to the principal axis after refraction [Fig. 9.19(b)]. Figures 9.19(a) and (b) illustrate these rules for a convex and a concave lens, respectively. You should practice drawing similar ray diagrams for different positions of the object with respect to the lens and also verify that the lens formula, Eq. (9.23), holds good for all cases.

Here again it must be remembered that each point on an object gives out infinite number of rays. All these rays will pass through the same image point after refraction at the lens.

Magnification (m) produced by a lens is defined, like that for a mirror, as the ratio of the size of the image to that of the object. Proceeding in the same way as for spherical mirrors, it is easily seen that for a lens

$$m=\frac{h^{\prime }}{h}=\frac{v}{u}$$

When we apply the sign convention, we see that, for erect (and virtual) image formed by a convex or concave lens, m is positive, while for an inverted (and real) image, m is negative.

9.5.3 Power of a lens

Power of a lens is a measure of the convergence or divergence, which a lens introduces in the light falling on it. Clearly, a lens of shorter focal length bends the incident light more, while converging it in case of a convex lens and diverging it in case of a concave lens. The power P of a lens is defined as the tangent of the angle by which it converges or diverges a beam of light parallel to the principal axis falling at unit distance from the optical centre (Fig. 9.20).

$$\tan \delta =\frac{h}{f};\text{ if }h=1,\tan \delta =\frac{1}{f}\text{ or }\delta =\frac{1}{f}$$ Thus,

$$P=\frac{1}{f}$$

The SI unit for power of a lens is dioptre (D): 1D = 1$m^{–1}$ . The power of a lens of focal length of 1 metre is one dioptre. Power of a lens is positive for a converging lens and negative for a diverging lens. Thus, when an optician prescribes a corrective lens of power + 2.5 D, the required lens is a convex lens of focal length + 40 cm. A lens of power of – 4.0 D means a concave lens of focal length – 25 cm

9.5.4 Combination of thin lenses in contact

Consider two lenses A and B of focal length $f_1$ and $f_2$ placed in contact with each other. Let the object be placed at a point O beyond the focus of the first lens A (Fig. 9.21). The first lens produces an image at $I_1$ . Since image $I_1$ is real, it serves as a virtual object for the second lens B, producing the final image at I. It must, however, be borne in mind that formation of image by the first lens is presumed only to facilitate determination of the position of the final image. In fact, the direction of rays emerging from the first lens gets modified in accordance with the angle at which they strike the second lens. Since the lenses are thin, we assume the optical centres of the lenses to be coincident. Let this central point be denoted by P.

For the image formed by the first lens A, we get $$\frac{1}{v_1}-\frac{1}{u}=\frac{1}{f_1}$$ For the image formed by the second lens B, we get $$\frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2}$$ Adding Eqs. (9.27) and (9.28), we get $$\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}$$ If the two lens-system is regarded as equivalent to a single lens of focal length f, we have $$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$ so that we get

$$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$$

The derivation is valid for any number of thin lenses in contact. If several thin lenses of focal length $f_1$ , $f_2$ , $f_3$ ,... are in contact, the effective focal length of their combination is given by

$$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}+\dots$$

In terms of power, $$P=P_1+P_2+P_3+\dots$$

where P is the net power of the lens combination. Note that the sum in Eq. (9.32) is an algebraic sum of individual powers, so some of the terms on the right side may be positive (for convex lenses) and some negative (for concave lenses). Combination of lenses helps to obtain diverging or converging lenses of desired magnification. It also enhances sharpness of the image. Since the image formed by the first lens becomes the object for the second, Eq. (9.25) implies that the total magnification m of the combination is a product of magnification ($m_1$ , $m_2$ , $m_3$ ,...) of individual lenses

$$m=m_1{m}_2{m}_3\dots$$

Such a system of combination of lenses is commonly used in designing lenses for cameras, microscopes, telescopes and other optical instruments.