 # Geometric Optics The Law of Reflection.

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Geometric Optics The Law of Reflection

Optics The Study of Light

Areas of Optics Geometric Optics Physical Optics Quantum Optics
Light as a ray. Physical Optics Light as a wave. Quantum Optics Light as a particle.

Reflection Reflection occurs when light bounces off a surface.
There are two types of reflection Specular reflection Off a shiny surface Diffuse reflection Off a rough surface

Mirrors are great reflectors
concave convex shiny dark shiny dark shiny dark + - + - + - Plane Mirror Spherical Mirrors

Light Rays Mathematical rays never bend
But light rays can, if they interact with materials!

Let’s take a closer look at a plane mirror
Incident ray Reflected ray + - normal A normal is a line that is perpendicular to the mirror. Plane Mirror

Ray tracing Ray tracing is a method of constructing an image using the model of light as a ray. We use ray tracing to construct optical images produced by mirrors and lenses. Ray tracing lets us describe what happens to the light as it interacts with a medium.

Law of Reflection Lab Hold a plane mirror upright on a sheet of graph paper that is on top of cardboard. Stick three pins through the graph paper on the same side of the mirror such that they appear to your eye to be in a straight line. (See the whiteboard for details) Draw a normal to the surface of the mirror, as well as the incident and reflected rays. Measure the angles of incidence and reflection. Repeat two or three times with different angles. Tabulate your angles of incidence and reflection. What can you say about the angles of incidence and reflection? TURN IN ONE DRAWING PER GROUP. Include each person’s name and period number.

Law of Reflection The angle of incidence of reflected light equals the angle of reflection. r = I Note that angles are measured relative to a normal to the mirror surface. shiny (+) dark (-) reflected ray r i normal incident ray light source plane mirror

Sample Problem A ray of light reflects from a plane mirror with an angle of incidence of 37o. If the mirror is rotated by an angle of 5o, through what angle is the reflected ray rotated?

Solution qr = 420 50 qi = 420 qr = 370 qi = 370
42o + 5o = 47o relative to horizontal 47o - 37o = 10o rotation of reflected ray

Sample Problem Standing 2.0 m in front of a small vertical mirror, you see the reflection of your belt buckle, which is 0.70 m below your eyes What is the vertical location of the mirror relative to the level of your eyes? If you move backward until you are 6.0 m from the mirror, will you still see the buckle, or will you see a point on your body that is above or below the buckle?

Solution eye mirror 0.35 m 0.70 m qr 2.0 m qi
The mirror must be 0.35 m below the level of your eye. If you move backward, you’ll still see the belt buckle, since the triangles will still be similar. buckle

The Plane Mirror and Optical Images
Wednesday, August 30, 2006 The Plane Mirror and Optical Images

Optical images Nature Orientation Size real (converging rays)
virtual (diverging rays) Orientation upright inverted Size true enlarged reduced

Ray tracing: plane mirror
Construct the image using two rays. + - Image -5 cm Extend reflected rays behind mirror. object 5 cm Reflected rays are diverging. Name the image: Virtual, upright, true size

Spherical mirrors shiny + - shiny + - concave convex
There are two types of spherical mirrors shiny + - shiny + - (where reflected rays go) (dark side) (where reflected rays go) (dark side) concave convex Focal length, f, is positive Focal length, f, is negative

Parts of a Spherical Concave Mirror
These are the main parts of a spherical concave mirror. The focal length is half of the radius of curvature. The focal length is positive for this type of mirror. R = 2f + - Center R Focus f Principle axis

Identification of the focus of a spherical concave mirror
Rays parallel to the principle axis all pass through the focus for a spherical concave mirror. + -

Ray tracing: spherical concave mirror
The three “principle rays” to construct an image for a spherical concave mirror are the p-ray, which travels parallel to the principle axis, then reflects through focus. the f-ray, which travels through focus, then reflects back parallel to the principle axis. the c-ray, which travels through center, then reflects back through center. You must draw two of the three principle rays to construct an image.

Ray tracing: spherical concave mirror
Construct the image for an object located outside the center of curvature. It is only necessary to draw 2 of the three principle rays! p f C F Real, Inverted, Reduced Image

Ray tracing: spherical concave mirror
Construct the image for an object located at the center of curvature. Name the image. C F Real, Inverted, True Image

Ray tracing: spherical concave mirror
Construct the image for an object located between the center of curvature and the focus. Name the image. C F Real, Inverted, Enlarged Image

Ray tracing: spherical concave mirror
Construct the image for an object located at the focus. F C No image is formed.

Ray tracing: spherical concave mirror
Construct the image for an object located inside the focus. Name the image. C F Virtual, Upright, Enlarged Image

Problem Construct 2 ray diagrams to illustrate what happens to the size of the image as an object is brought nearer to a spherical concave mirror when the object outside the focus. Repeat part a) for an object which is brought nearer to the mirror but is inside the focus.

Solution a) The image becomes larger when you move the object closer.

Solution b) The image becomes smaller when you move the object closer.

Mirror equation #1 1/si + 1/so = 1/f si: image distance
so: object distance f: focal length

Mirror equation # 2 M = hi/ho = -si/so si: image distance
so: object distancem hi: image height ho: object height M: magnification

Sample Problem A spherical concave mirror, focal length 20 cm, has a 5-cm high object placed 30 cm from it. Draw a ray diagram and construct the image. Name the image

Sample Problem A spherical concave mirror, focal length 20 cm, has a 5-cm high object placed 30 cm from it. Use the mirror equations to calculate the position of image the magnification the size of image

Parts of a Spherical Convex Mirror
These are the main parts of a spherical convex mirror. The focal length is half of the radius of curvature, and both are on the dark side of the mirror. The focal length is negative for this type of mirror. + - Principle axis Focus Center

Ray tracing: spherical convex mirror
Construct the image for an object located outside a spherical convex mirror. Name the image. F C Virtual, Upright, Reduced Image

Problem Construct 2 ray diagrams to illustrate what happens to the size of the image as an object is brought nearer to a spherical convex mirror.

Problem A spherical concave mirror, focal length 10 cm, has a 2-cm high object placed 5 cm from it. Draw a ray diagram and construct the image.

Problem A spherical concave mirror, focal length 10 cm, has a 2-cm high object placed 5 cm from it. Use the mirror equations to calculate the position of image the magnification the size of image Name the image

Problem A spherical convex mirror, focal length 15 cm, has a 4-cm high object placed 10 cm from it. Draw a ray diagram and construct the image.

Problem A spherical convex mirror, focal length 15 cm, has a 4-cm high object placed 10 cm from it. Use the mirror equations to calculate the position of image the magnification the size of image Name the image

Summary Concave vs convex mirrors
Image is real when object is outside focus Image is virtual when object is inside focus Focal length f is positive Convex Image is always virtual Focal length f is negative

Definition: Refraction
Refraction is the movement of light from one medium into another medium. Refraction cause a change in speed of light as it moves from one medium to another. Refraction can cause bending of the light at the interface between media.

n = c/v n = Index of Refraction speed of light in vacuum
speed of light in medium n = c/v n =

When light slows down… …it bends. Let’s take a look at a simulation.
URL:

n1 n2 Snell’s Law n1sin 1 = n2sin 2 1 2 angle of incidence
angle of refraction

n1 n2 n1 < n2 1 2 When n1 < n2 1 > 2
When the index of refraction increases, light bends toward the normal. n1 < n2 1 n1 n2 2 When n1 < n2 1 > 2

n1 n2 n1 > n2 1 2 When n1 > n2 1 < 2
When the index of refraction decreases, light bends away from the normal. n1 > n2 1 n1 2 n2 When n1 > n2 1 < 2

Problem Light enters an oil from the air at an angle of 50o with the normal, and the refracted beam makes an angle of 33o with the normal. Draw this situation. Calculate the index of refraction of the oil. Calculate the speed of light in the oil

Problem Light enters water from a layer of oil at an angle of 50o with the normal. The oil has a refractive index of 1.65, and the water has a refractive index of 1.33. Draw this situation. Calculate the angle of refraction. Calculate the speed of light in the oil, and in the water

Problem Light enters a prism as shown, and passes through the prism.
Complete the path of the light through the prism, and show the angle it will make when it leaves the prism. If the refractive index of the glass is 1.55, calculate the angle of refraction when it leaves the prism. How would the answer to b) change if the prism were immersed in water? air 30o glass 60o

Problem Light enters a prism made of air from glass. 30o glass air 60o
Complete the path of the light through the prism, and show the angle it will make when it leaves the prism. If the refractive index of the glass is 1.55, calculate the angle of refraction when it leaves the prism. 30o glass air 60o

Dispersion The separation of white light into colors due to different refractive indices for different wavelengths is called dispersion. Dispersion is often called the prism effect.

Dispersion Which color of light has the greatest refractive index?

Critical Angle of Incidence
The smallest angle of incidence for which light cannot leave a medium is called the critical angle of incidence. If light passes into a medium with a greater refractive index than the original medium, it bends away from the normal and the angle of refraction is greater than the angle of incidence. If the angle of refraction is > 90o, the light cannot leave the medium.

Critical Angle of Incidence
This drawing reminds us that when light refracts from a medium with a larger n into one with a smaller n, it bends away from the normal. n2 n1 > n2

Critical Angle of Incidence
Ray reflects instead of refracting. c n1 This shows light hitting a boundary at the critical angle of incidence, where the angle of refraction is 90o. No refraction occurs! r = 90o n2 n1 > n2

Critical Angle of Incidence
Ray reflects instead of refracting. c n1 Instead of refraction, total internal reflection occurs when the angle of incidence exceeds the critical angle. r = 90o n2 n1 > n2

Calculating Critical Angle
n1sin(1) = n2sin(90o) n1sin(c) = n2sin(90o) sin(c) = n2/ n1 c = sin-1(n2/n1)

Problem (separate sheet)
What is the critical angle of incidence for a gemstone with refractive index 2.45 if it is in air? If you immerse the gemstone in water (refractive index 1.33), what does this do to the critical angle of incidence?

Lenses - + - + There are two types of lenses. converging diverging
(where refracted rays go) - + (where refracted rays go) - + Thicker in middle Thinner in middle Focal length, f, is positive Focal length, f, is negative converging diverging

Lens ray tracing Ray tracing is also used for lenses.
We use the same principle rays we used for mirrors. the p-ray, which travels parallel to the principle axis, then refracts through focus. the f-ray, which travels through focus, then refracts parallel to the principle axis. the c-ray, which travels through center and continues without bending. You must draw 2 of the 3 principle rays.

Lens Equations We use the same equations we used for mirrors
1/si + 1/so = 1/f M = hi/ho = -si/so

Identification of the focus
All rays parallel to the principle axis refract through the focus of a converging lens. F

Ray tracing: converging lens
+ - Construct the image for an object located outside 2F. It is only necessary to draw 2 of the three principle rays! p c f 2F F C F 2F Real, Inverted, Reduced Image

Ray tracing: converging lens
+ - Construct the image for an object located at 2F. 2F F C F 2F Real, Inverted, True Image

Ray tracing: converging lens
+ - Construct the image for an object located between F and 2F. 2F F C F Real, Inverted, Enlarged Image

Ray tracing: converging lens
+ - Construct the image for an object located at the focus. 2F F C F No image

Ray tracing: converging lens
+ - Construct the image for an object located inside the focus. F C F Virtual, Upright, Enlarged Image

For converging lenses f is positive so is positive
si is positive for real images and negative for virtual images M is negative for real images and positive for virtual images hi is negative for real images and positive for virtual images

Diverging lens + - Construct the image for an object located in front of a diverging lens. F C F Virtual, Upright, Reduced Image

For diverging lenses f is negative so is positive si is negative
M is positive and < 1 hi is positive and < ho

Problem A converging lens, focal length 20 cm, has a 5-cm high object placed 30 cm from it. Draw a ray diagram and construct the image. Use the lens equations to calculate the position of image the magnification the size of image Name the image

Solution a) and c) + - a) F C F c) Real, Inverted, Enlarged

Solution b) 1/si + 1/so = 1/f 1/si + 1/30 = 1/20
si = 60 cm M = - si/so = -60/30 = -2 M = hi/ ho hi = M ho= (-2)5 = -10 cm

Problem A converging lens, focal length 10 cm, has a 2-cm high object placed 5 cm from it. Draw a ray diagram and construct the image. Use the lens equations to calculate the position of image the magnification the size of image Name the image

Solution a) and c) + - a) F C F c) Virtual, Upright, Enlarged

Solution b) 1/si + 1/so = 1/f 1/si + 1/5 = 1/10
si = -10 cm M = - si/so = -(-10)/5 = 2 M = hi/ ho hi = M ho= (2)2 = 4 cm

Problem A diverging lens, focal length -15 cm, has a 4-cm high object placed 10 cm from it. Draw a ray diagram and construct the image. Use the lens equations to calculate the position of image the magnification the size of image Name the image

Solution a) and c) + - a) F C F c) Virtual, Upright, Reduced

Solution b) 1/si + 1/so = 1/f 1/si + 1/10 = 1/(-15)
= -5/30 = -1/6 si = -6 cm M = - si/so = -(-6)/10 = 0.6 M = hi/ ho hi = M ho= (0.6)(4) = 2.4 cm