DEMONSTRATE UNDERSTANDING OF WAVES AS 2.3 LIGHT 4 WEEKS WAVES 4 WEEKS

 Light info  Lenses/mirrors  Concave/convex/plane  Reflection/refraction  Incident/reflected rays  Ray diagrams – how to draw and components needed  Laws of reflection

LIGHT  Type of electromagnetic radiation  Oscillating electric and magnetic fields  Travel together at a speed of 3.0x 10 8 ms ¯¹ (speed of light)  Form of energy  Appears to travel in straight lines

 A lens is an optical device with perfect or approximate axial symmetry which transmits and refracts (bends) light, converging or diverging the beam.  A mirror is an object with at least one reflective surface. A plane mirror has a flat surface. Curved mirrors are also used, to produce magnified or diminished images or focus light

 Concave:  curves inwards  Convex:  curves outwards  Plane:  flat surface

 Rays and wave fronts can generally be used to represent light when the light is interacting with objects. We will mainly use rays to analyse how light interacts with mirrors and lenses.

Reflection  Reflection from a mirror: Incident ray Normal Reflected ray Angle of incidence Angle of reflection Mirror

Incident and reflected rays  The ray of light approaching the mirror is known as the incident ray  The ray of light that leaves the mirror is known as the reflected ray  At the point of incidence where the ray strikes the mirror, a line can be drawn perpendicular to the surface of the mirror. This line is known as a normal line  The normal line divides the angle between the incident ray and the reflected ray into two equal angles.  The angle between the incident ray and the normal is known as the angle of incidence.  The angle between the reflected ray and the normal is known as the angle of reflection.

Laws of reflection The incident ray, reflected ray and the normal all lie in the same plane 2. The angle of incidence is equal to the angle of reflection

Ray diagrams  Are diagrams drawn to scale and used to determine the details of an image produced by either a lens or mirror  You need to know how to figure out the following details of the image produced  Position: how far along the principal axis, may be either behind or in front of the mirror  Size: cm/height  Nature: upright/inverted, virtual or real, magnified or diminished

 Focal point: the point through which all parallel light rays are reflected through  Centre of curvature: centre point circle  Radius of curvature: distance between pole and centre of curvature (same as radius of circle)  The focal length of a curved mirror is half the radius of curvature  Focal point: the point through which all parallel light rays are reflected through  Centre of curvature: centre point circle  Radius of curvature: distance between pole and centre of curvature (same as radius of circle)  The focal length of a curved mirror is half the radius of curvature

 A mirror is drawn as a curved line and/or a straight mirror line to make ray diagrams easier. Dashed lines or shading on the non-reflective side.

 Draw a ray diagram of:  A concave mirror, with principal axis, focal point of 4cm and centre of curvature of 8cm.  Draw an object (arrow) of 2cm height 5cm from the pole.

 1. Look at the diagram below.  Which one of the angles (A, B, C, or D) is the angle of incidence? ______  Which one of the angles is the angle of reflection? ______

 2. A ray of light is incident towards a plane mirror at an angle of 30-degrees with the mirror surface. What will be the angle of reflection?

 3. A ray of light is approaching a set of three mirrors as shown in the diagram. The light ray is approaching the first mirror at an angle of 45-degrees with the mirror surface. Trace the path of the light ray as it bounces off the mirror. Continue tracing the ray until it finally exits from the mirror system. How many times will the ray reflect before it finally exits?

 4. You can observe the image of the sun in the windows of distant buildings near the time that the sun is rising or setting. However, the image of the sun is not seen in the windows of distant building during midday. Use the diagram below to explain.

Formula  : is the image distance  : is the object distance  Magnification factor:

ConvergingDiverging Concave Convex Focal lengthf>0 +ve focal length f<0 -ve focal length

 The terms concave and convex don’t tell you whether the mirror or lens is converging or diverging and therefore whether the focal length is positive or negative.  Need to remember the table.

Other variable  Object distance: for single lens or mirror is always positive.  Image distance: i is positive for a real image  i is negative for a virtual image  Image location: (reflected rays) where the outgoing light rays or their tracebacks converge  Real: reflected rays converge  Virtual: traceback rays converge

 Real image: same side as the reflected light positive image distance.  Virtual: opposite side (to outgoing light) negative image distance.  Magnification: >1 magnified  <1 diminished

 So we know that if our mirror or lens is converging or diverging-this tells us about the focal point  Real or virtual tells us about the image distance  Magnification tells us whether the image is smaller or larger than the object.  We can then use what we know to help confirm whether out ray diagram is correct.

CONCAVE MIRRORS  Used to converge light and focus it on a point  Law 1: Any light ray that is parallel to the principal axis will be reflected through the focal point  Law 2: Any light ray that passes through the focal point will be reflected back parallel to the principal axis

Converging mirror  Real vs Virtual image  Real means that the image is formed by converging reflected light rays.  If the object gets too close to the mirror (closer than the focal point), the reflected rays diverge. The human eye can still see this image but it is virtual and cannot be projected onto a screen.  Ray diagram-if the rays diverge, continue on the reflected rays behind the mirror to the point where they converge

Review mirrors/lenses/ray diagrams Ray diagrams cont….. Use formula to find unknowns LIGHT……

1. An incoming ray of light parallel to the principal axis will be ___________________________ 2. Any light ray that passes through the focal point will be ________________________________ 3. How is a virtual image formed?

 What is this and what will it do to the incoming light rays?  Draw a convex mirror-what will it do to the incoming light rays?

 An incoming ray of light parallel to the principal axis will be …  reflected back through the focus  Any light ray that passes through the focal point will be ….  reflected back parallel to the principal axis  How is a virtual image formed?  The reflected light rays converge and so tracebacks need to be continued to form a virtual image.

 What is this and what will it do to the incoming light rays?  Concave lens that will diverge the light rays.  Draw a convex mirror-what will it do to the incoming light rays?  Diverge the light rays.

Formula example….. FIND THE FOCAL POINT OF A MIRROR IF THE OBJECT DISTANCE IS 9CM AND THE IMAGE DISTANCE IS 18CM

CONCAVE MIRRORS...  Don’t need to use negative number for magnification equation  If you need to work out the size of the image, you need to work out the magnification and then multiply the object size by this magnification.

Laws Formulas Ray diagrams CONVEX MIRRORS

 Law 1: any incoming light ray that is parallel to the principal axis is reflected back so it is aligned with the far focal point  Law 2: any incoming light ray directed towards the focal point is reflected back parallel to the principal axis

CONVEX MIRRORS  Image is always virtual i.e. Formed from traceback rays.  And always behind the mirror.  Image is formed between the focus and the mirror.  Same formulas apply

 The focal point is always negative as it is behind the mirror.

CONVEX MIRRORS  ALWAYS:  Located behind the convex mirror  A virtual image  An upright image  Reduced in size

 Practice:  Using a ray diagram find the image location and size.  Convex mirror, focal length of 8cm, 3cm high, and placed 4cm away from the mirror.

Convex Mirror Practice Ray diagrams Pg 108 Formula questions Pg 109

 Real image video

CONVEX MIRRORS…….  Name 2 features of an image formed by convex mirror.  What does a convex mirror do to light rays?  Why is the focal point negative?  What is the formula for magnification?

CONCAVE PRAC 2...  Investigate the formula  Set up the mirror as you did for the last experiment.  Repeat step 2 to find the COC and F points for your mirror.  Then measure 6-10 different values for object & image distance in a table.  Complete the table as it is set up in your workbook.  What happens if you try distances less than the focal point?  Write this up with conclusion about what you have found-you need to be able to explain your processes.

REFRACTION OF LIGHT

REFRACTION:  When a light ray refracts, it bends.  The light rays will only bend at the boundary between 2 different substances. Prior to and after that the light rays will travel in a straight line.  The refraction is caused by the light changing speed as it enters a new medium.  The amount of refraction depends on the refractive index of the substance.

 n= speed of light in a vacuum speed of light in medium

SNELLS LAW.....  The refractive index of a medium is given by Snell’s law  If a ray of light passes across the boundary from a material in which it travels fast into a material in which travels slower, then the light ray will bend towards the normal line.  If a ray of light passes across the boundary from a material in which it travels slowly into a material in which travels faster, then the light ray will bend away from the normal line.

 h/Optics/Refraction/Refraction.html h/Optics/Refraction/Refraction.html

A ray of light strikes a glass block (n=1.5) at 40 deg to the normal. Find the angle through which the ray is deviated. Draw a diagram to model this situation. 11. Assume  = 1 sin (40) 1.50 22. θ = 25.4 deg 33. The angle of deviation will therefore be = 14.6 deg.

Refractive index/Snells Law Questions Pages