Index of Refraction

The Index of Refraction Experimental Fact Light slows down when traveling through a medium. The Index of Refraction n, of the medium is defined as the ratio of the speed of light in vacuum, c, to the speed of light in the medium, v: or v = (c/n)

Refraction: Snell’s Law Light changes direction when crossing a boundary from one medium to another. This is called refraction. The angle the outgoing ray makes with the normal is called the angle of refraction. Figure 32-21. Refraction. (a) Light refracted when passing from air (n1) into water (n2): n2 > n1. (b) Light refracted when passing from water (n1) into air (n2): n2 < n1.

Refraction The phenomenon that causes objects half submerged in water to look odd. Figure 32-22. Ray diagram showing why a person’s legs look shorter when standing in waist-deep water: the path of light traveling from the bather’s foot to the observer’s eye bends at the water’s surface, and our brain interprets the light as having traveled in a straight line, from higher up (dashed line).

The angle of refraction depends on the indices of refraction, and is given by Snell’s Law: An experimentally verified law. But, comes directly from Maxwell’s theory of electromagnetic radiation.

Refraction through flat glass Example Refraction through flat glass Light traveling in air strikes a flat piece of uniformly thick glass at an incident angle of θ1 = 60° , as shown. Index of refraction of glass n1 = 1.50 . Calculate (a) The angle of refraction θA in the glass. (b) The angle θB at which the ray emerges from the glass. Solution: a. Applying Snell’s law gives sin θA = 0.577, or θA = 35.3°. b. Snell’s law gives sin θB = 0.866, or θB = 60°. The outgoing ray is parallel to the incoming ray.

Apparent depth of a pool Example Apparent depth of a pool A swimmer has dropped her goggles to the bottom of a pool at the shallow end, marked as d = 1.0 m deep. But the goggles don’t look that deep. (a) Why? (b) How deep do the goggles appear to be when you look straight down into the water? That is, find d' in the figure. Solution: The ray diagram appears in Figure 32-25. Refraction causes the goggles to appear to be less deep than they actually are. Snell’s law plus a small-angle approximation (sin θ ≈ tan θ ≈ θ) gives d’ ≈ d/n1 = 0.75 m.

Figure 35.10 (a) The wave under refraction model. (b) Light incident on the Lucite block refracts both when it enters the block and when it leaves the block. Fig. 35-10a, p. 1017

Visible Spectrum and Dispersion The visible spectrum contains the full range of wavelengths of light that are visible to the human eye. Figure 32-26. The spectrum of visible light, showing the range of wavelengths for the various colors as seen in air. Many colors, such as brown, do not appear in the spectrum; they are made from a mixture of wavelengths.

The index of refraction of many transparent materials (such as glass & water) varies slightly with wavelength. This is how prisms & water droplets create rainbows from sunlight. Figure 32-28. Index of refraction as a function of wavelength for various transparent solids. Figure 32-29. White light dispersed by a prism into the visible spectrum.

λn = (v/f) = (c/nf) = (λ/n) This spreading of light into the full spectrum is called dispersion. Recall that for a wave of frequency f, wavelength λ, & speed v, a general relation is v = f λ. Figure 32-20. (a) Ray diagram explaining how a rainbow (b) is formed. λn = (v/f) = (c/nf) = (λ/n)

Conceptual Example Observed color of light under water Light’s color depends on it’s wavelength. For example, an object emitting λ = 650 nm light in air looks red. But this is true only in air. If this same object is observed when under water, it still looks red. But the wavelength in water λn = 650 nm/1.33 = 489 nm. Light with wavelength λ = 489 nm would appear blue in air. Can you explain why the light appears red rather than blue when observed under water? Solution: Evidently our eyes respond to the frequency of the light (which does not change) rather than its wavelength. We refer to wavelengths as they are much more easily measured.

Total Internal Reflection; Fiber Optics If light passes into a medium with a smaller index of refraction, the angle of refraction is larger. There is a critical incidence angle, θ1 = θC, for which the angle of refraction will be θ2 = 90°. This is called the critical angle. From Snell’s Law this is given by: But, for θ2 = 90°, sinθ2 = 1 & θ1 = θC. So

If the angle of incidence is larger than this, no transmission occurs If the angle of incidence is larger than this, no transmission occurs. This is called Total Internal Reflection. Figure 32-31. Since n2 < n1, light rays are totally internally reflected if the incident angle θ1 > θc, as for ray L. If θ1 < θc, as for rays I and J, only a part of the light is reflected, and the rest is refracted.

Conceptual Example View up from under water Describe what a person would see who looked up at the world from beneath the perfectly smooth surface of a lake or swimming pool. Figure 32-32. (a) Light rays, and (b) view looking upward from beneath the water (the surface of the water must be very smooth). Example 32–11. Solution: The critical angle for an air-water interface is 49°, so the person will see the upwards view compressed into a 49° circle.

Binoculars often use total internal reflection; this gives true 100% reflection, which even the best mirror cannot do. Figure 32-33. Total internal reflection of light by prisms in binoculars.

Fiber Optics Optical fibers also depend on total internal reflection; they are therefore able to transmit light signals with very small losses. Figure 32-34. Light reflected totally at the interior surface of a glass or transparent plastic fiber.

Refraction at a Spherical Surface Rays from a single point will be focused by a convex spherical interface with a medium of larger index of refraction to a single point, as long as the angles are not too large. Figure 32-37. Diagram for showing that all paraxial rays from O focus at the same point I (n2 > n1).

Geometry gives the relationship between the indices of refraction, the object distance, the image distance, & the radius of curvature:

For a concave spherical interface, the rays will diverge from a virtual image. Figure 32-38. Rays from O refracted by a concave surface form a virtual image (n2 > n1). Per our conventions, R < 0, di < 0, do > 0.

Example: Apparent depth II. A person looks vertically down into a 1.0-m-deep pool. How deep does the water appear to be? Solution: Using equation 32-8 gives di = -0.75 m.

Example: A spherical “lens.” A point source of light is placed at a distance of 25.0 cm from the center of a glass sphere of radius 10.0 cm. Find the image of the source. Solution: Use equation 32-8 twice (once at each interface); the first image is at -90.0 cm, and the final image is at 28 cm.

Summary of Chapter Light paths are called rays. Index of refraction. Angle of reflection equals angle of incidence. Plane mirror: The image is virtual, upright, & the same size as the object. Spherical mirror can be concave or convex. Focal length of the mirror Mirror equation: Magnification:

Light passes through real images. Law of refraction (Snell’s law): Light does not pass through virtual images. Law of refraction (Snell’s law): Total internal reflection occurs when angle of incidence is greater than the critical angle: