Ray Optics, Mirrors, Lenses, Image and Optical Instruments Chapter 22, 23 Ray Optics, Mirrors, Lenses, Image and Optical Instruments

A Brief History of Light 1000 AD It was proposed that light consisted of tiny particles Newton Used this particle model to explain reflection and refraction Huygens 1670 Explained many properties of light by proposing light was wave-like

A Brief History of Light, cont Young 1801 Strong support for wave theory by showing interference Maxwell 1865 Electromagnetic waves travel at the speed of light

A Brief History of Light, final Planck EM radiation is quantized Implies particles Explained light spectrum emitted by hot objects Einstein Particle nature of light Explained the photoelectric effect

in the absence of medium. C = f l C: speed of light C = 3 x 108 m/s in vacuum EM wave can travel in the absence of medium. In other medium, the speed of light is smaller than in vacuum.

3 x 108 = f l nano = 10-9 http://laxmi.nuc.ucla.edu:8248/M248_99/iphysics/spectrum.gif

How do we see an object? Detection of light directly emitted by object Detection of light reflected by object (most common)

Ray Optics – Using a Ray Approximation Light travels in a straight-line path in a homogeneous medium until it encounters a boundary between two different media The ray approximation is used to represent beams of light A ray of light is an imaginary line drawn along the direction of travel of the light beams

The Index of Refraction Speed of light c= 3x 108 m/s in vacuum. Speed of light is different (smaller) in other media The index of refraction, n, of a medium can be defined

Index of Refraction, cont For a vacuum, n = 1 For other media, n > 1 n is a unitless ratio Normal air: 1.0003 Water: 1.33 Flint glass: 1.66 Diamond 2.42

Reflection of Light A ray of light, the incident ray, travels in a medium When it encounters a boundary with a second medium, part of the incident ray is reflected back into the first medium This means it is directed backward into the first medium

Specular Reflection Specular reflection is reflection from a smooth surface The reflected rays are parallel to each other All reflection in this text is assumed to be specular

Diffuse Reflection Diffuse reflection is reflection from a rough surface The reflected rays travel in a variety of directions Diffuse reflection makes the road easy to see at night

Law of Reflection The normal is a line perpendicular to the surface It is at the point where the incident ray strikes the surface The incident ray makes an angle of θ1 with the normal The reflected ray makes an angle of θ1’ with the normal

Law of Reflection, cont The angle of reflection is equal to the angle of incidence θ1= θ1’

Reflection and Mirrors q1 = q1 q1 1 Law of reflection Specular Reflection Diffuse Reflection

Image forms at the point where the light rays converge. When we talk about an image, start from an ideal point light source. Every object can be constructed as a collection of point light sources. VIRTUAL IMAGE |q| p Image forms at the point where the light rays converge. When real light rays converge  Real Image When imaginary extension of L.R. converge  Virtual Image Only real image can be viewed on screen placed at the spot.

For plane mirror: p = |q| IMAGE VIRTUAL For plane mirror: p = |q| How about left-right? Let’s check? p: object distance q: image distance

Parallel light rays: your point light source is very far away. Spherical Mirror R: radius of curvature focal Point f: focal length = R/2 Optical axis concave convex Parallel light rays: your point light source is very far away. Focal point: (i) Parallel incident rays converge after reflection (ii) image of a far away point light source forms (iii) On the optical axis

Spherical Aberration Reflected rays do not converge: Not well-defined focal point not clear image Spherical Aberration f = R/2 holds strictly for a very narrow beam. Parabolic mirror can fix this problem.

Case 1: p > R p f q P > q Real Image

Case 2: p = R p = q Real Image

Case 3: f < p < R p < q Real Image

Case 4: p = f q = infinite

Case 5: p < f q < 0 Virtual Image

For a small object, f = R/2 (spherical mirror) Mirror Equation 1/p + 1/q = 1/f For a small object, f = R/2 (spherical mirror) 1/p + 1/q = 2/R Alert!! Be careful with the sign!! Negative means that it is inside the mirror!! p can never be negative (why?) negative q means the image is formed inside the mirror VIRTUAL How about f?

Focal point inside the mirror For a concave mirror: f > 0 Focal point inside the mirror f < 0 1/p + 1/q = 1/f < 0 : q should be negative.

All images formed by a convex mirror are VIRTUAL. 1/p + 1/q = 1/f < 0 : q should be negative. All images formed by a convex mirror are VIRTUAL. Magnification, M = -q/p Negative M means that the image is upside-down. For real images, q > 0 and M < 0 (upside-down).

Example: An object is 25 cm in front of a concave spherical mirror of radius 80 cm. Determine the position and characteristics of the image. 1/p + 1/q = 1/f f = R/2 = 40 cm Object is at the center: p = 25 cm 1/q = 1/40 – 1/25 = -0.015 q = -66.7 cm < 0 (Virtual Image, 66.7 cm behind mirror) M = -q/p = -(-66.7)/25 = 2.7 Erect, 2.7 times the size of the object

Example: What kind of spherical mirror must be used, and what must be its radius, in order to give an erect image 1/5 as Large as an object placed 15 cm in front of it? M = -q/p  -q/p=1/5 So q = -p/5 = -15/5 = -3 cm 1/p + 1/q = 1/f  1/15 - 1/3 = 1/f 1/f = (1-5)/15 f = -15/4 = -3.75 cm R = 2 f = -7.5 cm Convex

Example: Where should an object be placed with reference to a concave spherical mirror of radius 180 cm in order to form a Real image having half its size? M = -q/p  -q/p=-1/2 So q = p/2 f = R/2 = 90 cm 1/p + 1/q = 1/f  1/p + 2/p = 1/f 3/p=1/90 p = 270 cm

Refraction Details, 1 Light may refract into a material where its speed is lower The angle of refraction is less than the angle of incidence The ray bends toward the normal

Refraction Details, 2 Light may refract into a material where its speed is higher The angle of refraction is greater than the angle of incidence The ray bends away from the normal

All three beams (incident, reflected, and refracted) are in one plane. Snell’s Law All three beams (incident, reflected, and refracted) are in one plane. q1 q1 q2 n > 1 n1sinq1 = n2sinq2

q1 q1 q1> q2 q2 water

Total Internal Reflection Total internal reflection can occur when light attempts to move from a medium with a high index of refraction to one with a lower index of refraction Ray 5 shows internal reflection

Critical Angle A particular angle of incidence will result in an angle of refraction of 90° This angle of incidence is called the critical angle

Critical Angle, cont For angles of incidence greater than the critical angle, the beam is entirely reflected at the boundary This ray obeys the Law of Reflection at the boundary Total internal reflection occurs only when light attempts to move from a medium of higher index of refraction to a medium of lower index of refraction

Examples of critical angles (relative to vacuum) Substance N Critical angle Vacuum 1 90.0 Air 1.00029 88.6 Ice 1.31 49.8 Water 1.333 48.6 Ethyl Alcohol 1.36 47.3 Glycerine 1.473 42.8 Crown glass 1.52 41.1 Sodium chloride 1.54 40.5 Quartz 1.544 40.4 Heavy flint glass 1.65 37.3 Tooth enamel 1.655 37.2 Sapphire 1.77 34.4 Heaviest flint glass 1.89 31.9 Diamond 2.42 24.4  

qf = 2qc Fish vision qc = sin-1(1/1.33) = 49 How could fish survive from spear fishing? Fish vision qf = 2qc qc = sin-1(1/1.33) = 49

ncore nclad >

Lenses Converging lens Diverging lens

Thin Lenses A thin lens consists of a piece of glass or plastic, ground so that each of its two refracting surfaces is a segment of either a sphere or a plane Lenses are commonly used to form images by refraction in optical instruments

Thin Lens Shapes These are examples of converging lenses They have positive focal lengths They are thickest in the middle

More Thin Lens Shapes These are examples of diverging lenses They have negative focal lengths They are thickest at the edges

Glass lens (nG = 1.52)

Same shape lenses: the higher n, the shorter f The focal length of a lens is determined by the shape and material of the lens. Same shape lenses: the higher n, the shorter f Lenses with same n: the shorter radius of curvature, the shorter f Typical glass, n = 1.52 Polycarbonate, n = 1.59 (high index lens) Higher density plastic, n ≈ 1.7 (ultra-high index lens)

off the optical axis as a point light source. Rules for Images Trace principle rays considering one end of an object off the optical axis as a point light source. A ray passing through the focal point runs parallel to the optical axis after a lens. A ray coming through a lens in parallel to the optical axis passes through the focal point. A ray running on the optical axis remains on the optical axis. A ray that pass through the geometrical center of a lens will not be bent. Find a point where the principle rays or their imaginary extensions converge. That’s where the image of the point source.

two focal points: f1 and f2 Parallel rays: image at infinite!!

Virtual image Magnifying glass Virtual image

positive negative p q f M 1/p + 1/q = 1/f M = -q/p Lens equation and magnification 1/p + 1/q = 1/f M = -q/p This eq. is exactly the same as the mirror eq. Now let’s think about the sign. positive negative p real object virtual object (multiple lenses) q real image (opposite side of object) virtual image (same side of object) f for converging lens for diverging lens M erect image inverted image

two focal points: f1 and f2 Parallel beams: image at infinite!! 1/p + 1/q = 1/f 1/2f + 1/q = 1/f 1/q = 1/2f M = -q/p = -1 two focal points: f1 and f2 1/p + 1/q = 1/f 1/f + 1/q = 1/f 1/q = 0  q = infinite Parallel beams: image at infinite!!

Virtual image Magnifying glass 1/p + 1/q = 1/f 2/f + 1/q = 1/f 1/q = -1/f M = -(-f)/(f/2) = 2 Virtual image

real image (opposite side) positive f Example: A thin converging lens has a focal length of 20 cm. An object is placed 30 cm from the lens. Find the image Distance, the character of image, and magnification. f = 20, p = 30 1/q = 1/f – 1/p = 1/20 – 1/30 = 1/60 q = 60 real image (opposite side) M = -q/p = -60/30 = -2 < 0 inverted

Magnifier Consider small object held in front of eye Height y Makes an angle  at given distance from the eye Goal is to make object “appear bigger”: ' >  y 

Rays seen coming from here Magnifier Single converging lens Simple analysis: put eye right behind lens Put object at focal point and image at infinity Angular size of object is , bigger! Outgoing rays Rays seen coming from here y   f f Image at Infinity

(angular) Magnification One can show f must be in cm

Example Find angular magnification of lens with f = 5 cm

Combination of Thin Lenses, example

Telescope View Distant Objects (Angular) Magnification M=fobj/feye Increased Light Collection Large Telescopes use Mirrors