## Friday, January 9, 2015

### Sign Convention and Image Tracing of Light in Ray Optics

Sign convention

We shall follow certain sign convention in measuring the distances while we are solving the problems are dealing the concepts in Ray optics.

For the sign convention, pole is taken as the origin and the principal axis as the x-axis.
We shall assume that light is always coming from left to right in the given diagram.

All the distances are measured from the pole.

If the distance of the object is measured along the direction of the incident light, it shall be treated like a positive.

If the distance of the object is measured against the direction of incident light, then it is treated as negative.

The same rules are valid even when we are measuring the distance of the image.

The heights measured upward normal to the principal axis are treated as positive and the height measured in the downward direction is treated as negative.

Focal length and the radius of curvature for a convex mirror are treated as positive and for a concave mirror they are treated as negative. It is simply because these values, when measured from the pole appear along the same direction of the incident light as shown in the diagram below.

Image tracing

When a point object is placed before a spherical mirror, a point images formed. The point of intersection of the incident rays is called object and the point of intersection of reflected light rays is called image.

To measure object distance, image distance, focal length and radius of curvature we need to follow certain sign convention. If the reflected light rays intersect with each other, then at the point of intersection is the place a real images formed. If the reflected light rays diverged from the surface of the reflection, the image is a virtual image. It is going to form at the point from where these light rays are appearing like diverging.
Basing on the sign convention we can take their values as positive or negative as shown in the diagram below.

Problem and solution

A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its one end close to the pole is 20 cm away from the mirror. What is the length of the image formed in this case?

We have to solve this problem in  two different parts. Let us consider the rod as combination of two points where one is the starting point and other one is the ending point.

We can apply mirror formula for both the points with appropriate sign convention as shown below.

Radius of curvature is numerically double to the value of the focal length. Basing on the values of object distance, image distance and the focal length of a mirror, we can derive a relation for magnification. Magnification is simply defined as the ratio of heat of the image to the height of the object. It can be also defined as the ratio of image distance to the object distance.

Relation between them can be shown as

Problem and solution

An object is placed on the principal axis of a concave mirror of focal length 10 cm at a distance 8 cm from the pole. Find the position and the nature of the image?

While solving this problem, we shall apply the proper sign convention. Being the mirror is a concave mirror; its focal length is negative.

Object is placed before the concave mirror as shown. As the object distance is measured from the pole which is against the direction of the incident light Ray, it shall be treated as negative. We don’t know the value of the image location therefore we are not going to assign any specific sign to it. We will take it as it there in the formula and basing an answer will conclude that what is the location and the nature of the image is.
As for the formula it can be identified that the image distances +40 cm. It means that it is developed at the other end of the mirror and then only can be treated as positive. It means the image is a virtual image.

Problem and solution

The above diagram is having another problem with the solution.

At what distance from the concave mirror of focal and 25 cm boy shall stand so that his image as a high equal to half of its original height?

The magnification of the concave mirror is negative which means to say that object and image will never be along the same direction. If object distances positive image distances negative and vice versa. By applying the basic formula of magnification as ratio of image distance to the object distance and the mirror formula with can solve the problem as shown above.

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