**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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