Reflection and mirrors

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Presentation transcript:

Reflection and mirrors

Reflections from plane mirrors The first mirrors appeared about 4000 years ago when the Egyptians used polished copper or bronze plates for that purpose. It wasn’t until 1857 when Jean Foucault (Foo-co) developed a method of using silver nitrate to make mirrors. Modern mirrors are made by evaporating aluminum or silver to the surfaces of highly polished glass. The mirrors used in industry have to be manufactured in highly controlled, dust free environments. The mirror used in the Hubble Space Telescope was designed to be flat to 10 nanometers. There was a problem with the mirror (only discovered after it was in space) when it was discovered to be too flat by 2,200 nanometers.

This error in the grinding process led to a spherical aberration – much like being near sighted or far sighted. The light gets focused in the wrong place. The problems with the Hubble were corrected by 2 service missions by members of different shuttle missions. The Law of Reflection – The law of reflection can be summed up as:  incidence =  reflection This means that the angle of the incoming ray is identical to the angle of the outgoing ray.

Terms to be familiar with A mirror is a polished, opaque surface that reflects light in predictable ways. The incident ray is the ray approaching a mirror. The reflected ray is a ray reflected by the mirror. The point of incidence is the point where the incident ray strikes the mirror. The normal is an imaginary line drawn at a right angle (90○) to the mirror at the point of incidence. The angle of incidence is the angle between the incident ray and the normal (often given as  i ) The angle of reflection is the angle between the reflected ray and the normal (often given as  r ) Dashed lines will indicate the back of the mirror Regular (specular) reflection occurs on smooth surfaces. Diffuse reflection occurs on rough surfaces (like the chalk board) Reflex (spread) reflection occurs on surfaces that are a combination of smooth and textured surfaces.

Properties of plane-mirror images Note: A plane-mirror is a flat mirror. The characteristics of images produced by plane-mirrors include: The image is the same size as the object. The image is erect The image is virtual (it cannot be focused on screen but can be photographed) The image is laterally inverted (the left side of the object looks like the right side of the image.) The image is the same distance behind the mirror as the object is in front of the mirror.

Curved mirrors We will be looking at 2 types of curved mirrors: Concave Mirrors – In a concave mirror the reflective surface curves inward, and causes the parallel beams of light to converge (come together) at a focal point. It can also be called a converging mirror.

Convex mirror – In a convex mirror the reflective surface curves outward, causing parallel beams of light to diverge (spread out). It is important to note that the focal point on this mirror is virtual (it is located behind the mirror).

Components of curved mirrors Centre of Curvature (C) – the centre of the curved reflecting surface. Radius of Curvature (R) – any straight line drawn from the centre of curvature to the curved surface. Vertex (V) – the geometric centre of the curved surface. Principal Axis – straight line passing through the centre of curvature and the vertex. Focal Point (F)– the point where the light rays converge. The rays parallel to the axis reflect off a concave mirror and converge at the focal point.

The focal length of a curved mirror is found by taking the radius of the mirror and dividing it by 2. The radius of a concave mirror is 6 cm. The focal point would be located 3 cm from the vertex of the mirror. The type of image created by a converging mirror depends on where the object is located. If the object is located between C and F, then the image will be larger and inverted. This image will not be observable unless either the viewer is at exactly the right location (where the image is showing up) or the observer is look at the image on an opaque screen.

If the object is beyond c When the object is located at a location beyond the center of curvature, the image will always be located somewhere in between the center of curvature and the focal point. Regardless of exactly where the object is located, the image will be located in the specified region. It will be smaller and inverted.

When the object is located right at c When the object is located at the center of curvature, the image will also be located at the center of curvature. In this case, the image will be inverted (i.e., a right side up object results in an upside-down image). Object and image will be the same size.

The object is located at f When the object is located at the focal point, no image is formed.

The object is in front of F When the object is located at a location beyond the focal point, the image will always be located somewhere on the opposite side of the mirror. Regardless of exactly where in front of F the object is located, the image will always be located behind the mirror.

There are 2 different methods for finding image size, orientation, and magnification. The first is graphical. You draw a diagram of the mirror and the object. You need to know the centre of curvature (C) and the focal length (fl) (the focal length will give you the focal point). You also need to know the height of the object and where it is located. You will end up with a diagram similar to one of the previous slides. To draw the diagram you will need : A ray that is parallel to the principal axis and reflected back through F. A ray that passes through F and is reflected back parallel to the principal axis. A ray that passes through C and is reflected back along the same path.

Example : Object located beyond F (and C)

Mathematical method of locating the image This method tends to be much easier than drawing the image. The comparison of the size of the image to the size of the object is called magnification. It can also be the ratio of the image distance to the object distance. magnification = height of the image height of the object m = hi = -di ho do you can also say: hi = - di ho do

Focal length : + for converging (concave) mirror - for diverging (convex) mirror do is the object distance from the vertex This is always a positive value di is the image distance from the vertex This is always positive for a real image This is always negative for a virtual image A real image is an inverted image formed by converging rays. A virtual image is formed by diverging rays and is always behind the mirror. The mirror equation: 1/f = 1/di + 1/do (this is used to determine image position)

Example: Given: do = 30. 0 cm (located 30 Example: Given: do = 30.0 cm (located 30.0 cm from a converging mirror) f = 5.0 cm ho = 4.0 cm (assuming upright on the principal axis) Find: di and hi 1/di + 1/do = 1/f  1/di = 1/f – 1/do  1/5.0 cm – 1/30.0 cm because we are working with fractions, we need to find LCD. In this case the LCD is 30, so we need to multiply 1/5.0 cm by 6 to give us 6/30.0cm We can then re-write this as: 1/di = 6/30.0cm – 1/30.0 cm = 5 / 30.0cm Now take the reciprocal of both sides: di = 30.0cm / 5.0cm = 6.0 cm The image is 6.0 cm from the vertex.

Finding hi hi / h0 = - di / do hi = ho x (-di / d0 ) = (4 Finding hi hi / h0 = - di / do hi = ho x (-di / d0 ) = (4.0 cm) x ( - 6.0 cm / 30.0 cm) = -0.80 cm The negative value means the image is inverted and reduced in size.

Assignment Using 1/f = 1/ di + 1/do determine di if you have a concave mirror with a focal length of 16.0 cm and an object positioned 36.0 cm in front of the mirror. You have a concave mirror with a focal length of 7.0 cm and a centre of curvature of 10.0 cm. You have a 2.4 cm tall object located 16.0 cm from the mirror. Determine the height and position of the image by drawing a scale diagram. (remember you have 3 rays – one from the top of the object through C, one from the top of the object through f that strikes the mirror and comes back parallel, and one from the top of the object parallel to the axis that strikes the mirror and reflects back through f.)