2A.5.2 – Reflection of Large Objects

Up till now, we have mostly been dealing with point objects (i.e. the object is a single point). In this section, we will look at large objects (consisting of many point objects).

Locate a Large (Extended) Image in a Mirror

The method simply requires us to consider the extended object to be made up of 2 or more points.

If we wish to locate the mirror image of this arrow

we can image the arrow as being defined by its two end points (A and B).
Reflect A to find the position of A’. This will be the position of the arrow tip.
Reflect B to find the position of B’. This will be the position of the arrow tail.
We can join these two points
We have thus shown the reflected position of this extended object.

Ray Diagrams Showing an Extended Object

To see a large object in a mirror, the mirror must be large enough such that light rays from the furthest edges of the object can reflect off the mirror to get to the eye.

Hence, there will also be two light rays. However, instead of two light rays starting at the middle of the object, and ending at two different points on the eye, this situation has light rays starting at different points on the object (usually from two extreme ends of the object, e.g. top and bottom), but ending at the same point on the eye.

 

Example Size of Mirror Using Similar Triangle Method
A boy stands 3.0 m away from wall AD facing the wall  BC in a room that is 8.0 m by 5.0 m. Determine the minimum length of the mirror required to be mounted on wall BC so that the boy can see the full length of the image of the wall AD in the mirror.

First step is always to locate the position of the image.

A straight line from the boy to the edge of the image, A’, shows gives us the location of Z – one edge of the ends of the mirror.

Another straight line from the boy to the edge of the image, D’, shows gives us the location of Y – the other edge of the mirror. Thus a mirror stretching from Z to Y (having length L) will allow the boy to just see the whole reflection of the wall AD.

We could show the original paths of these incident rays.

Similar triangles allow us to determine the length of YZ, L.

Triangle xZY is similar to triangle xA’D’, thus:

L/5 = 5.0/13.0

L=1.9 m (2 s.f.)

 

 

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