Geometric Optics and Image Formation

12 Images Formed by Refraction

Learning Objectives

By the end of this section, you will be able to:

  • Describe image formation by a single refracting surface
  • Determine the location of an image and calculate its properties by using a ray diagram
  • Determine the location of an image and calculate its properties by using the equation for a single refracting surface

When rays of light propagate from one medium to another, these rays undergo refraction, which is when light waves are bent at the interface between two media. The refracting surface can form an image in a similar fashion to a reflecting surface, except that the law of refraction (Snell’s law) is at the heart of the process instead of the law of reflection.

Refraction at a Plane Interface—Apparent Depth

If you look at a straight rod partially submerged in water, it appears to bend at the surface ((Figure)). The reason behind this curious effect is that the image of the rod inside the water forms a little closer to the surface than the actual position of the rod, so it does not line up with the part of the rod that is above the water. The same phenomenon explains why a fish in water appears to be closer to the surface than it actually is.

Bending of a rod at a water-air interface. Point P on the rod appears to be at point Q, which is where the image of point P forms due to refraction at the air-water interface.

Figure depicts the side view of a rod dipped in water. A lighter line labeled image of rod is shown in such a way that it appears as if the rod is bent at the junction of air and water. Point P is on the rod and point Q is on the image of rod. A dotted line PQ is shown perpendicular to the surface of the water. Two rays originate from P, travel upwards to the surface of the water, bend at an angle and reach the eye of the observer. The back extensions of the bent rays seem to originate from point Q.

To study image formation as a result of refraction, consider the following questions:

  1. What happens to the rays of light when they enter or pass through a different medium?
  2. Do the refracted rays originating from a single point meet at some point or diverge away from each other?

To be concrete, we consider a simple system consisting of two media separated by a plane interface ((Figure)). The object is in one medium and the observer is in the other. For instance, when you look at a fish from above the water surface, the fish is in medium 1 (the water) with refractive index 1.33, and your eye is in medium 2 (the air) with refractive index 1.00, and the surface of the water is the interface. The depth that you “see” is the image height {h}_{\text{i}} and is called the apparent depth. The actual depth of the fish is the object height {h}_{\text{o}}.

Apparent depth due to refraction. The real object at point P creates an image at point Q. The image is not at the same depth as the object, so the observer sees the image at an “apparent depth.”

Figure shows the side view of some quantity of water. Point P lies within it. Two rays originate from point P, bend at the surface of the water and reach the eye of the observer. The back extensions of these refracted rays intersect at point Q. PQ is perpendicular to the surface of the water and intersects it at point O. Distance OP is labeled h subscript o and distance OQ is labeled h subscript i. The angle formed by the refracted ray with a line perpendicular to the surface of the water is labeled theta.

The apparent depth {h}_{\text{i}} depends on the angle at which you view the image. For a view from above (the so-called “normal” view), we can approximate the refraction angle \theta to be small, and replace sin \theta in Snell’s law by tan \theta. With this approximation, you can use the triangles \text{Δ}OPR and \text{Δ}OQR to show that the apparent depth is given by

{h}_{\text{i}}=\left(\frac{{n}_{2}}{{n}_{1}}\right){h}_{\text{o}}.

The derivation of this result is left as an exercise. Thus, a fish appears at 3/4 of the real depth when viewed from above.

Refraction at a Spherical Interface

Spherical shapes play an important role in optics primarily because high-quality spherical shapes are far easier to manufacture than other curved surfaces. To study refraction at a single spherical surface, we assume that the medium with the spherical surface at one end continues indefinitely (a “semi-infinite” medium).

Refraction at a convex surface

Consider a point source of light at point P in front of a convex surface made of glass (see (Figure)). Let R be the radius of curvature, {n}_{1} be the refractive index of the medium in which object point P is located, and {n}_{2} be the refractive index of the medium with the spherical surface. We want to know what happens as a result of refraction at this interface.

Refraction at a convex surface \left({n}_{2}>{n}_{1}\right).

Figure shows a section of a sphere. The refractive index of air is n subscript 1 and that of the sphere is n subscript 2. Centre of the sphere is C and radius is R. A ray originating from point P on the optical axis outside the sphere strikes the convex surface of the sphere and is refracted within it. It intersects the axis at point P prime within the sphere, on the other side of the center. A dotted line labeled normal to interface connects the center of the sphere to the point of incidence. It makes an angle phi with the optical axis. The incident and refracted rays make angles alpha and beta respectively with the optical axis and angles theta 1 and theta 2 respectively with the normal to interface.

Because of the symmetry involved, it is sufficient to examine rays in only one plane. The figure shows a ray of light that starts at the object point P, refracts at the interface, and goes through the image point {P}^{\prime }. We derive a formula relating the object distance {d}_{\text{o}}, the image distance {d}_{\text{i}}, and the radius of curvature R.

Applying Snell’s law to the ray emanating from point P gives {n}_{1}\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta }_{1}={n}_{2}\text{sin}\phantom{\rule{0.2em}{0ex}}{\theta }_{2}. We work in the small-angle approximation, so \text{sin}\phantom{\rule{0.2em}{0ex}}\theta \approx \theta and Snell’s law then takes the form

{n}_{1}{\theta }_{1}\approx {n}_{2}{\theta }_{2}.

From the geometry of the figure, we see that

{\theta }_{1}=\alpha +\varphi ,\phantom{\rule{0.6em}{0ex}}{\theta }_{2}=\varphi -\beta .

Inserting these expressions into Snell’s law gives

{n}_{1}\left(\alpha +\varphi \right)\approx {n}_{2}\left(\varphi -\beta \right).

Using the diagram, we calculate the tangent of the angles \alpha ,\beta ,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\varphi:

\text{tan}\phantom{\rule{0.2em}{0ex}}\alpha \approx \frac{h}{{d}_{\text{o}}},\phantom{\rule{0.6em}{0ex}}\text{tan}\phantom{\rule{0.2em}{0ex}}\beta \approx \frac{h}{{d}_{\text{i}}},\phantom{\rule{0.6em}{0ex}}\text{tan}\phantom{\rule{0.2em}{0ex}}\varphi \approx \frac{h}{R}.

Again using the small-angle approximation, we find that \text{tan}\phantom{\rule{0.2em}{0ex}}\theta \approx \theta, so the above relationships become

\alpha \approx \frac{h}{{d}_{\text{o}}},\phantom{\rule{0.6em}{0ex}}\beta \approx \frac{h}{{d}_{\text{i}}},\phantom{\rule{0.6em}{0ex}}\varphi \approx \frac{h}{R}.

Putting these angles into Snell’s law gives

{n}_{1}\left(\frac{h}{{d}_{\text{o}}}+\frac{h}{R}\right)={n}_{2}\left(\frac{h}{R}-\frac{h}{{d}_{\text{i}}}\right).

We can write this more conveniently as

\frac{{n}_{1}}{{d}_{\text{o}}}+\frac{{n}_{2}}{{d}_{\text{i}}}=\frac{{n}_{2}-{n}_{1}}{R}.

If the object is placed at a special point called the first focus, or the object focus {F}_{1}, then the image is formed at infinity, as shown in part (a) of (Figure).

(a) First focus (called the “object focus”) for refraction at a convex surface. (b) Second focus (called “image focus”) for refraction at a convex surface.

Figure a shows a section of a sphere and a point F1 outside it, on the optical axis. Rays originating from F1 strike the convex surface and are refracted within the sphere as parallel rays. The distance of F1 from the surface is f subscript 1. Figure b shows rays parallel to the optical axis striking the convex surface and being refracted. They converge at point F2 within the sphere. F2 lies on the optical axis between the surface and the center of the sphere. The distance of F2 from the surface is f subscript 2. In both figures the refractive index of air is n1 and that of the sphere is n2 greater than n1.

We can find the location {f}_{1} of the first focus {F}_{1} by setting {d}_{\text{i}}=\text{∞} in the preceding equation.

\frac{{n}_{1}}{{f}_{1}}+\frac{{n}_{2}}{\text{∞}}=\frac{{n}_{2}-{n}_{1}}{R}
{f}_{1}=\frac{{n}_{1}R}{{n}_{2}-{n}_{1}}

Similarly, we can define a second focus or image focus {F}_{2} where the image is formed for an object that is far away [part (b)]. The location of the second focus {F}_{2} is obtained from (Figure) by setting {d}_{\text{o}}=\text{∞}:

\frac{{n}_{1}}{\text{∞}}+\frac{{n}_{2}}{{f}_{2}}=\frac{{n}_{2}-{n}_{1}}{R}
{f}_{2}=\frac{{n}_{2}R}{{n}_{2}-{n}_{1}}.

Note that the object focus is at a different distance from the vertex than the image focus because {n}_{1}\ne {n}_{2}.

Sign convention for single refracting surfaces

Although we derived this equation for refraction at a convex surface, the same expression holds for a concave surface, provided we use the following sign convention:

  1. R>0 if surface is convex toward object; otherwise, R<0.
  2. {d}_{\text{i}}>0 if image is real and on opposite side from the object; otherwise, {d}_{\text{i}}<0.

Summary

This section explains how a single refracting interface forms images.

  • When an object is observed through a plane interface between two media, then it appears at an apparent distance {h}_{\text{i}} that differs from the actual distance {h}_{\text{o}}: {h}_{\text{i}}=\left({n}_{2}\text{/}{n}_{1}\right){h}_{\text{o}}.
  • An image is formed by the refraction of light at a spherical interface between two media of indices of refraction {n}_{1} and {n}_{2}.
  • Image distance depends on the radius of curvature of the interface, location of the object, and the indices of refraction of the media.

Conceptual Questions

Derive the formula for the apparent depth of a fish in a fish tank using Snell’s law.

Use a ruler and a protractor to find the image by refraction in the following cases. Assume an air-glass interface. Use a refractive index of 1 for air and of 1.5 for glass. (Hint: Use Snell’s law at the interface.)

(a) A point object located on the axis of a concave interface located at a point within the focal length from the vertex.

(b) A point object located on the axis of a concave interface located at a point farther than the focal length from the vertex.

(c) A point object located on the axis of a convex interface located at a point within the focal length from the vertex.

(d) A point object located on the axis of a convex interface located at a point farther than the focal length from the vertex.

(e) Repeat (a)–(d) for a point object off the axis.

answers may vary

Problems

An object is located in air 30 cm from the vertex of a concave surface made of glass with a radius of curvature 10 cm. Where does the image by refraction form and what is its magnification? Use {n}_{\text{air}}=1 and {n}_{\text{glass}}=1.5.

An object is located in air 30 cm from the vertex of a convex surface made of glass with a radius of curvature 80 cm. Where does the image by refraction form and what is its magnification?

{d}_{\text{i}}=-55\phantom{\rule{0.2em}{0ex}}\text{cm};m=+1.8

An object is located in water 15 cm from the vertex of a concave surface made of glass with a radius of curvature 10 cm. Where does the image by refraction form and what is its magnification? Use {n}_{\text{water}}=4\text{/}3 and {n}_{\text{glass}}=1.5.

An object is located in water 30 cm from the vertex of a convex surface made of Plexiglas with a radius of curvature of 80 cm. Where does the image form by refraction and what is its magnification? {n}_{\text{water}}=4\text{/}3 and {n}_{\text{Plexiglas}}=1.65.

{d}_{\text{i}}=-41\phantom{\rule{0.2em}{0ex}}\text{cm,}\phantom{\rule{0.2em}{0ex}}m=1.4

An object is located in air 5 cm from the vertex of a concave surface made of glass with a radius of curvature 20 cm. Where does the image form by refraction and what is its magnification? Use {n}_{\text{air}}=1 and {n}_{\text{glass}}=1.5.

Derive the spherical interface equation for refraction at a concave surface. (Hint: Follow the derivation in the text for the convex surface.)

proof

Glossary

apparent depth
depth at which an object is perceived to be located with respect to an interface between two media
first focus or object focus
object located at this point will result in an image created at infinity on the opposite side of a spherical interface between two media
second focus or image focus
for a converging interface, the point where a bundle of parallel rays refracting at a spherical interface; for a diverging interface, the point at which the backward continuation of the refracted rays will converge between two media will focus

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