## 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:

- The object is always placed towards the left of the mirror. Therefore, the incident light rays fall on the mirror from the left.
- All the distances parallel to the principal axis are measured from the pole of the mirror.
- All the distances measured to the right of the origin, along + x-axis, are taken as positive.
- All the distances measured to the left of the origin, along – x-axis, are taken as negative.
- All the distances measured perpendicular to the principal axis above it, along the +y-axis, are taken to be positive.
- 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 = – 6**0 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 = 1**0 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**.

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