Lenses and the Refraction of Light

26.1 The Index of Refraction Light travels through a vacuum at at c Light travels through materials such as water or glass at a speed less than c. Light will bend, or refract, if it enters a medium with a different speed. It may reflect as well. The amount of bending is quantified by the index of refraction. The index of refraction of a material is the ratio of the speed of light in a vacuum to the speed of light in the material. As a ratio, it is unitless:

26.2 Snell’s Law and the Refraction of Light SNELL’S LAW OF REFRACTION When light travels from a material with one index of refraction to a material with a different index of refraction, the angle of incidence is related to the angle of refraction by where n1 is the index of refraction in the incident ray medium, and n2 is that index of refraction in the refracting ray medium. There is also a reflected ray at an angle of θ1. A mirror surface does not refract, it only reflects.

Determining the Angle of Refraction A light ray strikes an air/water surface at an angle of 46 degrees with respect to the normal. Find the angle of refraction when the direction of the ray is (a) from air to water and (b) from water to air. From air to water From water to air n1 = 1 n1 = 1.33 θ1 = 46° θ1 = 46° n2 = 1.33 n2 = 1 Find θ2 Find θ2

26.2 Snell’s Law and the Refraction of Light When a ray is transmitted into a material with a higher index of refraction it bends to make a smaller angle with the normal. (b) When a ray is transmitted into a material with a lower index of refraction it bends to make a larger angle with the normal.

26.2 Snell’s Law and the Refraction of Light APPARENT DEPTH – when an object is in a medium with a different index of refraction that that of the observer, the observer will see a virtual image of the object which is at a different location than the object Apparent depth, observer directly above object Where n1 is for the medium of the incident ray coming from the object and n2 is for the medium of the refracted ray

Workbook exercises

26.2 Snell’s Law and the Refraction of Light Conceptual Example 4 On the Inside Looking Out A swimmer is under water and looking up at the surface. Someone holds a coin in the air, directly above the swimmer’s eyes. To the swimmer, the coin appears to be at a certain height above the water. Is the apparent height of the coin: Greater than its actual height Less than its actual height Equal to its actual height

26.2 Snell’s Law and the Refraction of Light THE DISPLACEMENT OF LIGHT BY A SLAB OF MATERIAL Incident ray and emergent ray (doubly refracted) are equal.

We can’t compare n1 to n3 without knowing n2. A light ray travels from medium 1 to medium 3 as shown. For these media, n3 = n1. n3 > n1. n3 < n1. We can’t compare n1 to n3 without knowing n2. Answer: B

26.3 Total Internal Reflection When light passes from a medium of larger refractive index into one of smaller refractive index, the refracted ray bends away from the normal. At some point, the incident angle will be such that the reflected angle is 90° from the normal, i.e. no part of the ray will refract out of the 2nd medium. That incident angle is the incident angle, θc. Recall sin 90° = 1 Critical angle

26.3 Total Internal Reflection Fiber Optics The glass core of an optical fiber has an index of refraction of 1.60. The cladding has an index of refraction of 1.48. What is the maximum angle a light ray coming from the air can make with the core wall if it is to remain inside the fiber?

26.3 Total Internal Reflection Fiber Optics The glass core of an optical fiber has an index of refraction of 1.60. The cladding has an index of refraction of 1.48. What is the maximum angle a light ray coming from the air can make with the core wall if it is to remain inside the fiber? Known Find ncore = 1.60 θ1 nclad = 1.48 n1 sinθ1 = n2 sin θ2 Law of refraction θ2 = 90 – θc Geometry – see illustration sin θc = nclad /ncore Total internal reflection

26.3 Total Internal Reflection Fiber Optics The glass core of an optical fiber has an index of refraction of 1.60. The cladding has an index of refraction of 1.48. What is the maximum angle a light ray coming from the air can make with the core wall if it is to remain inside the fiber? Known Find ncore = 1.60 θ1 nclad = 1.48 nair θ1 = ncore θ2 Law of refraction θ2 = 90 – θc Geometry – see illustration sin θc = nclad /ncore Total internal reflection θ1 =37.44 degrees.

26.5 The Dispersion of Light: Prisms and Rainbows The net effect of a prism is to change the direction of a light ray. Light rays corresponding to different colors bends by different amounts.

26.5 The Dispersion of Light: Prisms and Rainbows A narrow beam of sunlight is incident on a slab of crown glass at 45 degrees. Find the angle between the violet ray and the red ray in the glass.

26.5 The Dispersion of Light: Prisms and Rainbows A narrow beam of sunlight is incident on a slab of crown glass at 45 degrees. Find the angle between the violet ray and the red ray in the glass. Answer: 0.35 degrees

26.5 The Dispersion of Light: Prisms and Rainbows

26.6 Lenses Mirrors reflect light according to the Law of Reflection and create a real or virtual image of the light source. Lenses refract light according to Snell’s Law of Refraction and create a real or virtual image of the light source. Converging and diverging lenses are analogous to concave and convex mirrors.

26.6 Lenses With a converging lens, paraxial rays that are parallel to the principal axis converge to the focal point of the lens on the opposite side from the object.

26.7 The Formation of Images by Lenses Ray diagrams for converging lenses: Ray 1 is parallel when it leaves the object and is refracted so that it passes through the focal point on the other side of the lens. Ray 2 passes through the focal point on the same side as the object and is refracted so that it is parallel on the other side of the lens. Ray 3 passes though the principal axis in the middle of the lens and continues on in a straight line.

26.7 The Formation of Images by Lenses There are 3 different types of images formed by a converging lens, depending on the relationship between the object and the focal point of the lens.

Object more than twice the focal length from the lens In this example, when the object is placed further than twice the focal length from the lens, the real image is inverted and smaller than the object. When using a digital camera, the film is replaced by a charge coupling device (CCD), a semi-conductor diode that acts much like a solar cell and changes light into electrical current.

Object further away than focal length When the object is placed between F and 2F, the real image is inverted and larger than the object. To see the real image from a projector on the screen, you have to be between the lens and the screen and look at the reflection from the screen

26.7 The Formation of Images by Lenses IMAGE FORMATION BY A CONVERGING LENS (MAGNIFYING GLASS) 2 When the object is placed between F and the lens, the virtual image is upright and larger than the object. Now you look through the lens, but you don’t see the object itself, but rather it’s virtual image. Note the lens is not high enough for #2 , but I drew it in.

26.6 Lenses With a diverging lens, paraxial rays that are parallel to the principal axis appear to originate from the focal point on the same side as the object.

26.7 The Formation of Images by Lenses Ray diagrams for diverging lenses: Ray 1 is parallel when it leaves the object and is refracted so that it’s continuation passes through the focal point on the same side of the lens as the object. Ray 2 is refracted so that it is parallel and its continuation passes through the focal point on the opposite side of the lens. Ray 3 passes though the principal axis in the middle of the lens and continues on in a straight line.

26.7 The Formation of Images by Lenses IMAGE FORMATION BY A DIVERGING LENS 2 A diverging lens always forms an upright, virtual, reduced image. Ray 2 leaves the object and travels toward the focal point on the other side of the lens.

The image will be as it was, but much dimmer. A lens produces a sharply-focused, inverted image on a screen. What will you see on the screen if the lens is removed? The image will be as it was, but much dimmer. The image will be right-side-up and sharp. The image will be right-side-up and blurry. The image will be inverted and blurry. There will be no image at all. STT23.4

26.8 The Thin-Lens Equation and the Magnification Equation

26.8 The Thin-Lens Equation and the Magnification Equation Summary of Sign Conventions for Lenses di is + for a real image formed on the opposite side of the lens as the object. di is - for a virtual image formed on the same side of the lens as the object.

A converging lens A 20 cm high object is located 30.0 cm to the left of a converging lens whose focal length is 50.0 cm. What is the nature of the image? real, inverted and reduced real, inverted and enlarged virtual, upright and enlarged virtual, upright and reduced.

A converging lens A 20 cm high object is located 30.0 cm to the left of a converging lens whose focal length is 50.0 cm. What is the a) position and b)magnification of the image? Solve for di and m

Converging lens From the thin-lens equation, we obtain: The negative answer shows it is on the same side of the lens as the object. The magnification equation gives the magnification to be: The positive number shows it is upright.

A converging lens A 20 cm high object is located 30.0 cm to the left of a converging lens whose focal length is 50.0 cm. What is the a) position and b)magnification of the image? Solve for di and m

The cornea and the lens act to bend the incoming light waves. 26.10 The Human Eye Anatomy of the Eye The cornea and the lens act to bend the incoming light waves. The lens only contributes about 20-25% of the refraction, but its function is important, as it provides the fine-tuning mechanism. The retina acts as the film, or CCD and receives the image, which it then sends, via the optic nerve, to the brain, which interprets the image.

When the ciliary muscles relax, the lens becomes less curved. 26.10 The Human Eye OPTICS When the ciliary muscles relax, the lens becomes less curved. The radius of curvature, and therefore the focal length increases. The eye can focus on more distant objects. Of course there is a limit to how far away an object can be that can be focused at the retina, even for people with good vision.

When the ciliary muscles become tensed, the lens becomes more convex. 26.10 The Human Eye OPTICS When the ciliary muscles become tensed, the lens becomes more convex. The radius of curvature, and therefore the focal length, decreases. This allows the eye to focus on closer objects. Of course there is a limit to this close focus.

26.10 The Human Eye NEARSIGHTEDNESS For some people, the lens cannot become “relaxed” enough to focus far-away objects on the retina. Instead, the eye focuses the image in front of the retina. For the purposes of this discussion, “a distant” object is considered to be infinitely far away (d0 = ∞).

26.10 The Human Eye NEARSIGHTEDNESS A divergent lens in front of the eye creates a virtual image in front of the object, close enough for the nearsighted eye to see. The virtual image acts as a virtual object, from which the eye makes a real image on the retina.

The lens creates an image of the close object behind the retina. 26.10 The Human Eye FARSIGHTEDNESS The lenses of some people cannot tense up enough to focus on a close object. The lens creates an image of the close object behind the retina. This happens to all people as they age !! Now the objective is to create a virtual image further away than the object.

26.10 The Human Eye FARSIGHTEDNESS The lens creates a virtual image of the close object at the near point of the farsighted eye.

Example 12 Eyeglasses for the Nearsighted Person 26.10 The Human Eye Example 12 Eyeglasses for the Nearsighted Person A nearsighted person has a far point located only 521 cm from the eye. Assuming that eyeglasses are to be worn 2 cm in front of the eye, find the focal length needed for the diverging lens of the glasses so the person can see “distant” objects.

26.10 The Human Eye d0 = ∞ di = (521-2)= 519 cm Find f

THE REFRACTIVE POWER OF A LENS – THE DIOPTER 26.10 The Human Eye THE REFRACTIVE POWER OF A LENS – THE DIOPTER Optometrists who prescribe correctional lenses and the opticians who make the lenses do not specify the focal length. Instead they use the concept of refractive power.

26.11 Angular Magnification and the Magnifying Glass The size of the image on the retina determines how large an object appears to be.

26.11 Angular Magnification and the Magnifying Glass

26.11 Angular Magnification and the Magnifying Glass Example 14 A Penny and the Moon Compare the angular size of a penny held at arms length with that of the moon. Penny Moon

26.11 Angular Magnification and the Magnifying Glass of a magnifying glass

26.12 The Compound Microscope To increase the angular magnification beyond that possible with a magnifying glass, an additional converging lens can be included to “premagnify” the object. Angular magnification of a compound microscope

Angular magnification of an astronomical telescope 26.13 The Telescope Angular magnification of an astronomical telescope