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## Key Concepts

• New cartesian sign convention
• Mirror formula
• Solving problems using the mirror formula

### Introduction

When dealing with the reflection of light by spherical mirrors mathematically, a set of sign conventions is followed, called the New Cartesian Sign Convention. According to this convention, the pole of a spherical mirror is taken as the origin and the principal axis is taken as the x-axis of the cartesian coordinate system. It is used to solve the numerical problems on the reflection by spherical mirrors.

## Explanation

### Rules of the new cartesian sign convention:

The new cartesian sign convention says that:

1. The object is always placed towards the left of the mirror. Therefore, the incident light rays fall on the mirror from the left.
2. All the distances parallel to the principal axis are measured from the pole of the mirror.
3. All the distances measured to the right of the origin, along + x-axis, are taken as positive.
4. All the distances measured to the left of the origin, along – x-axis, are taken as negative.
5. All the distances measured perpendicular to the principal axis above it, along the +y-axis, are taken to be positive.
6. All the distances measured perpendicular to the principal axis below it, along the –y-axis, are taken to be negative.

For a concave mirror,

The focal length (f) is negative,

The radius of curvature (R) is negative,

For a convex mirror,

The focal length (f) is positive, and

The radius of curvature (R) is positive.

### The mirror formula

For a spherical mirror,

• The distance of the object from the pole is called the object distance and is denoted by u
• The distance of the image from the pole is called the image distance and is denoted by v

The mirror formula is a relationship between the object distance (u), image distance (v) and the focal length (f) of a spherical mirror.

The mirror formula is given by,

𝟏𝒇 = 𝟏𝒗 + 𝟏𝒖

This formula is valid for all the spherical mirrors in all situations. The new cartesian sign convention should be used before substituting the values in the expression.

### Numerical problems

1. A car is at a distance of 8 m from the rear-view mirror of focal length 5 m of a taxi. Identify the type of mirror and also find out the position and nature of the image formed.

Solution:

The mirror is a convex mirror, as they are used as rear-view mirrors.

Given that,

The object distance, u = 8 m

The image distance, v = ?

The focal length of the mirror, f = 5 m

On using the sign convention,

The object distance, u = – 8 m

And the focal length = + 5 m

From the mirror formula we have,

𝟏/𝒇 = 𝟏/𝒗 + 𝟏/𝒖

Or,

1/5 = 1/v + 1/−8

Or,

1/v = 1/8 + 1/5

Or,

1/v = 1/8+1/5 =  13/40

Or, v = 40/13 = 0.325 m

Therefore, the image of the object is formed at a distance of 0.325 m behind the mirror. The image is virtual and upright

2. An object is placed at a distance of 20 m from a concave mirror of focal length 15 m. At what distance should a screen be placed in order to obtain a sharp image of the object? What is the nature of the image formed?

Solution:

Given that,

The object distance, u = 20 m

The image distance, v = ?

The focal length of the mirror, f = 15 m

On using the sign convention,

The object distance, u = – 20 m

And the focal length, f = – 15 m

From the mirror formula we have,

1/f = 1/v + 1/u

Or,

1/−15 = 1/v + 1/−20

Or,

1/v =1/−15 − 1/−20 = 1/−15+ 1/20

Or,

1/v = 1/20−1/15 =  – 1/60

Or, v = – 60/1 = – 60 m

Therefore, the image of the object is formed at a distance of 60 m behind the mirror. As the image is formed in front of the mirror, it is real and inverted

3. An object is placed at some distance from a concave mirror of focal length 5 m. A screen placed at a distance of 10 m gives a sharp image of the object. What is the distance at which the object is placed from the mirror? What is the nature of the image?

Solution:

Given that,

The object distance, u = ?

The image distance, v = 10 m

The focal length of the mirror, f = 5 m

On using the sign convention,

The image distance, v = – 10 m

And the focal length, f = – 5 m

From the mirror formula we have,

1/f = 1/v + 1/u

Or,

1/−5 = 1/−10 + 1/u

Or,

1/u = 1/−5 − 1/−10 = 1/−5+ 1/10

Or,

1/v = 1/10 − 1/5 =  -1/10

Or, v = -10/1 = 10 m

Therefore, the object is located at a distance of 10 m in front of the mirror. The image is real and inverted

Alternative solution:

Given that,

The image distance, v = 10 m

The focal length of the mirror, f = 5 m

This means that the image is formed at a distance twice that of the focal length i.e., at the center of curvature as R = v = 2f.

For an object located at C, the image is formed at C itself.

Thus, the object is placed at a distance of 10 m in front of the mirror.

The image is real and inverted

## Summary

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