 # Reflection and Refraction. Reflection  Reflection occurs when light bounces off a surface.  There are two types of reflection – Specular reflection.

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Reflection and Refraction

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

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

Types of mirrors  Plane mirrors  Spherical mirrors

Law of Reflection  The angle of incidence of reflected light equals the angle of reflection.  θ R = θ I  Note that angles are measured relative to a normal to the mirror surface.

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

Sample Problem  A ray of light reflects from a plane mirror with an angle of incidence of 37 o. If the mirror is rotated by an angle of 5 o, through what angle is the reflected ray rotated relative to the horizontal axis? 5o5o 37 o 32 o

Ray tracing: plane mirror  Construct the image using two rays. 5 cm object -5 cm image Plus side Minus side Upright (virtual) Same size Reflected rays Are diverging Extend reflected rays behind mirror

The focal point is where the rays intersect! Going from concave to convex the focal point stays on the concave side of the mirror! Rays parallel to the principal axis all pass through the focus for a Spherical concave mirror.

The focal length is always ½ of the radius! r f r is the radius of curvature C is the center of the sphere F is the focal length of the mirror

θ Law of Reflection still works even on a curved surface (just like a rough surface) θ θ

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

Construct the image for an object located outside the center of curvature. It is only necessary to draw 2 of the three principal rays! p ray c ray F ray Real image, inverted image, reduced image

Construct the image for an object located at the center of curvature. Name the image. p ray F ray Real image, inverted image, true image

(this solution is completed on the next slide) p ray f ray c ray

This tells us the image is inverted (upside down), larger (than the object), and real (on the same side of the mirror as the object). p ray f ray c ray Real, Inverted, Enlarged Image

p ray f ray c ray Real, Inverted, Enlarged Image

Construct the image for an object located at the focus. p ray c ray No image

Construct the image for an object located inside the focus. Name the image. p ray c ray Virtual, Upright, Enlarged Image f ray

Object is at “C” Object and image meet! (As object moves toward mirror, image moves away) (remember to right-click anywhere to pause)

Object is between “C” and “F”

Object is between “F” and mirror

Side-by-side comparison (remember to right-click anywhere to pause)

Like some make-up mirrors… (between “F” and mirror, image is ALWAYS enlarged, virtual, and erect.)

What’s the difference between a REAL and a VIRTUAL image?

A REAL image is on the SAME SIDE of the mirror as the object making the image. The Mirage demo creates a real image. Usually you need to project the REAL image onto a screen so you can see it. Like Alice in Wonderland objects not in the mirror (in Wonderland) are REAL or in the REAL world.

What’s the difference between a REAL and a VIRTUAL image? A VIRTUAL image we trace the rays back to a point where the rays appear to diverge. The image appears to be on the opposite side of the reflective surfac

Mirror equations  1/s i + 1/s o = 1/f – s i : image distance – s o : object distance – f: focal length  M = h i /h o = -s i /s o –h i : image height – h o : object height – M: magnification

Sign conventions  Focal length (f) – Positive for CONCAVE mirrors – Negative for CONVEX mirrors  Magnification (M) – Positive for UPRIGHT images – Negative for INVERTED images – ENLARGED when M > 1 – REDUCED when M < 1  Image Distance – s i is POSITIVE for real images – s i is NEGATIVE for virtual images

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

Refraction The bending of light rays

Refraction Formulas  Index of Refraction n = index of refraction v = speed of light in material c = speed of light in a vacuum

Formulas  Snell’s Law n i sin  i = n r sin  r  i = angle of incidence n i = index of refraction for incident medium  r = angle of refraction n r = index of refraction for refracting medium

The speed of light in plastic is 2.00 x 10 8 m/s. What is the index of refraction of plastic? n = ? v = 2.00 x 10 8 m/s c = 3.00 x 10 8 m/s n = 3.00 x 10 8 m/s / 2.00 x 10 8 m/s n = 1.5 Sample Problem

Notes – Refraction/Snell’s Law When light passes from one medium into another, o part of the incident light is reflected at the boundary and o the remainder passes into the new medium -if the ray enters the new medium at an angle (other than perpendicular), the ray bends as it enters.

A ray of light traveling through air is incident upon a sheet of crown glass at an angle of 30 o. What is the angle of refraction? n i sin  i = n r sin  r n i = 1.00  i = 30 o n r = 1.52(1.00)(sin 30 o ) = (1.52)(sin  r )  r = ??0.500 = sin  r 1.52  r = 19.2 o Sample Problem

When light passes from one material into a second material where the index of refraction is less (say, from water into air) light bends away from the normal. At a particular incident angle, the angle of refraction will be 90 o, and the refracted ray will skim the surface. The angle of incidence at which this occurs is called the critical angle. Critical Angle

Sample Problem  Find the critical angle for diamond. n i = 2.42 (diamond) n i sin  i = n r sin  r  c = ? n r = 1.00 (vacuum)  r = 90 o  r = 90 o (2.42)(sin  c ) = (1.00)(sin 90 o )  c = 24.4 o

incident ray Normal (90 o ) θiθi θrθr θiθi θrθr

The arrangement shown at the left is thicker in the middle and converges the light. The arrangement at the right, however, is thinner in the middle than at the edges; it diverges light

30-2 Converging, because the rays intersect (converge)

30-2 Diverging, because the rays move away from each other (diverge)

Convex Lenses: The lens below is a convex lens – also known as a converging lens.

The lens below is a concave lens, also known as a diverging lens.

The thin lens equations are the same as the mirror equations. However, there are different sign conventions that go along with using the equation for lenses. The Thin Lens Equation

For converging lenses  f is positive  d o is positive  d i is positive for real images an negative for virtual images  M is negative for real images and positive for virtual images  h i is negative for real images and positive for virtual images

For diverging lenses  f is negative  d o is positive  d i is negative  M is positive and < 1  h i is positive and < h o

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