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11.4: Images Formed by Refraction

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    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. 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.

    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.
    Figure \(\PageIndex{1}\): 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.

    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 \(\PageIndex{2}\)). 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 \(h_i\) and is called the apparent depth. The actual depth of the fish is the object height \(h_o\).

    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.
    Figure \(\PageIndex{2}\): 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.”

    The apparent depth hi 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 \(θ\) to be small, and replace \(\sin θ\) in Snell’s law by \(\tan θ\). With this approximation, you can use the triangles \(ΔOPR\) and \(ΔOQR\) to show that the apparent depth is given by

    \[h_i= \left(\dfrac{n_2}{n_1}\right)h_o. \nonumber \]

    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 (Figure \(\PageIndex{3}\)). Let \(R\) be the radius of curvature, n1 be the refractive index of the medium in which object point P is located, and n2 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.

    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.
    Figure \(\PageIndex{3}\): Refraction at a convex surface (\(n_2>n_1\)).

    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 \(P′\). We derive a formula relating the object distance \(d_o\), the image distance \(d_i\), and the radius of curvature \(R\).

    Applying Snell’s law to the ray emanating from point \(P\) gives

    \[n_1\sin θ_1=n_2 \sin θ_2. \nonumber \]

    Within the small-angle approximation

    \[\sin θ≈θ, \nonumber \]

    Snell’s law then takes the form

    \[n_1θ_1≈n_2θ_2. \label{eq8} \]

    From the geometry of Figure \(\PageIndex{3}\), we see that

    \[θ_1=α+ϕ, \nonumber \]

    \[θ_2=ϕ−β. \nonumber \]

    Inserting both expressions into Equation \ref{eq8} gives

    \[n_1(α+ϕ)≈n_2(ϕ−β). \label{eq10} \]

    Using Figure \(\PageIndex{3}\), we calculate the tangent of the angles \(α\), \(β\), and \(ϕ\):

    • \(\tan α≈\dfrac{h}{d_o}\)
    • \(\tan β≈\dfrac{h}{d_i}\)
    • \(\tan ϕ≈\dfrac{h}{R}\)

    Again using the small-angle approximation, we find that \(\tan θ≈ θ\), so the above relationships become

    • \(α≈\dfrac{h}{d_o}\)
    • \(~β≈\dfrac{h}{d_i}\)
    • \(~ϕ≈\dfrac{h}{R}.\)

    Putting these angles into Equation \ref{eq10} gives

    \[n_1\left(\dfrac{h}{d_o}+\dfrac{h}{R}\right)=n_2 \left(\dfrac{h}{R}−\dfrac{h}{d_i}\right). \nonumber \]

    We can write this more conveniently as

    \[\dfrac{n_1}{d_o}+\dfrac{n_2}{d_i}=\dfrac{n_2−n_1}{R}.  \]

    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 Figure \(\PageIndex{4a}\).

    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.
    Figure \(\PageIndex{4}\): (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.

    We can find the location \(f_1\) of the first focus \(F_1\) by setting \(d_i=\infty\) in Equation \ref{eq20}.

    \[ \begin{align} \dfrac{n_1}{f_1}+\dfrac{n_2}{\infty} &=\dfrac{n_2−n_1}{R} \\[4pt] f_1 &=\dfrac{n_1R}{n_2−n_1} \end{align} \nonumber \]

    Similarly, we can define a second focus or image focus \(F_2\) where the image is formed for an object that is far away (Figure \(\PageIndex{4b}\)). The location of the second focus \(F_2\) is obtained from Equation \ref{eq20} by setting \(d_0=\infty\):

    \[ \begin{align} \dfrac{n_1}{\infty}+\dfrac{n_2}{f_2}=\dfrac{n_2−n_1}{R} \\[4pt] f_2=\dfrac{n_2R}{n_2−n_1}. \end{align} \nonumber \]

    Note that the object focus is at a different distance from the vertex than the image focus because \(n_1≠n_2\).

    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. \(d_i>0\) if image is real and on opposite side from the object; otherwise, \(d_i<0\).

    This page titled 11.4: Images Formed by Refraction is shared under a CC BY license and was authored, remixed, and/or curated by OpenStax.