Optics and Waves

Question 1: Discuss the geometry of refraction and reflection, including Snell’s law and total internal reflection.

Answer 1: When light is transmitted from one medium to another, its direction may be altered upon entering the new medium. This is known as refraction. The degree to which the light is refracted depends on the index of refraction, n, for each medium. The index of refraction is a ratio of the speed of light in a vacuum to the speed of light in the medium in question, n = c/vm. Since light can never travel faster than it does in a vacuum, the index of refraction is always greater than one. Snell’s law gives the equation for determining the angle of refraction: n1sin(?1) = n2sin(?2), where n is the index of refraction for each medium, and ? is the angle the light makes with the normal vector on each side of the interface between the two media.We will examine a special case by trying to determine the angle of refraction for light traveling from a medium with n1 = 3 to another medium with n2 = 1.5. The light makes an angle ?1 = 35° with the normal. Using Snell’s law, we find that sin(?2) = 1.15. Since this is not mathematically possible, we conclude that the light cannot be refracted. This case is known as total internal reflection. When light travels from a more dense medium to a less dense medium, there is a minimum angle of incidence, beyond which all light will be reflected. This critical angle is ?1 = sin-1(n2/n1). Fiber-optic cables make use of this phenomenon to ensure that the signal is fully reflected internally when it veers into the outer walls of the fiber.

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Question 2: Discuss thin lenses.

Answer 2: A lens is an optical device that redirects light to either converge or diverge in specific geometric patterns. Whether the lens converges or diverges is dependent on the lens being convex or concave, respectively. The particular angle of redirection is dictated by the lens’s focal length. For a converging lens, this is the distance from the lens that parallel rays entering from the opposite side would intersect. For a diverging lens, it is the distance from the lens that parallel rays entering the lens would intersect if they were reverse extrapolated. However, the focal length of a diverging lens is always considered to be negative. A thin lens is a lens whose focal length is much greater than its thickness. By making this assumption, we can derive many helpful relations.

Question 3: Discuss images, differentiating between real and virtual images.

Answer 3: In optics, an object’s image is what is seen when the object is viewed through a lens. The location of an object’s image is related to the lens’s focal length by the equation 1/do + 1/di = 1/f, where f is the focal length, and do and di are the distance of the object and its image from the lens, respectively. A positive di indicates that the image is on the opposite side of the lens from the object. If the lens is a magnifying lens, the height of the object may be different from that of its image, and may even be inverted. The object’s magnification, m, can be found as m = -di/do. The value for the magnification can then be used to relate the object’s height to that of its image: m = yi/yo. Note that if the magnification is negative, then the image has been inverted.Images may be either real or virtual. Real images are formed by light rays passing through the image location, while virtual images are only perceived by reverse extrapolating refracted light rays. Diverging lenses cannot create real images, only virtual ones. Real images are always on the opposite side of a converging lens from the object and are always inverted.

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