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The study of Physics Topics involves the exploration of matter, energy, and the forces that govern the universe.

## Derivation of Mirror Formula and Linear Magnification

The distance of an object from the pole of a mirror is known as object distance. Object distance is denoted by the letter u. The distance of image from the pole of a mirror is known as image distance. Image distance is denoted by the letter v. The distance of focus from the pole of a mirror is known as focal length.

Focal length is denoted by the letter /. There is a relationship between the object distance, image distance and focal length of a spherical mirror (concave mirror or convex mirror). This relationship is given by the mirror formula.

A formula which gives the relationship between image distance (v), object distance (u) and focal length (f) of a spherical mirror is known as the mirror formula. The mirror formula can be written as :

\(\frac{1}{\text { Image distance }}+\frac{1}{\text { Object distance }}\) = \(\frac{1}{\text { Focal length }}\)

or \(\frac{1}{v}+\frac{1}{u}\) = \(\frac{1}{f}\)

where v = distance of image from mirror

u = distance of object from mirror

f = focal length of the mirror

The mirror formula has three values in it. If any two values are known, the third value can be calculated. The known values should be put in this formula with their proper signs but no sign should be given to the unknown value to be calculated. Its proper sign will come by itself by calculations.

The proper signs of the known values of u, v or f can be found by using New Cartesian Sign Convention for spherical mirrors. Another point to be noted is that the mirror formula can be applied to both type of spherical mirrors : concave mirrors as well as convex mirrors. By using this mirror formula, we can find out the position and nature of the images formed by concave mirrors and convex mirrors. Before we do that, we will discuss the ‘magnification’ produced by spherical mirrors.

### Linear Magnification Produced by Mirrors

The size of image formed by a spherical mirror depends on the position of the object from the mirror. The image formed by a spherical mirror can be bigger than the object, equal to the object or smaller than the object. The size of the image relative to the object is given by the linear magnification. This is discussed.

The ratio of the height of image to the height of object is known as linear magnification. That is,

Magnification = \(\frac{\text { height of image }}{\text { height of object }}\)

or m = \(\frac{h_2}{h_1}\)

where m = magnification

h_{2} = height of image

and h_{1} = height of object

In our ray-diagrams, the object is always placed above the principal axis, so the height (h_{1}) of the object will always be positive. We also know that a virtual image is always formed above the principal axis, therefore, the height (h_{2}) of a virtual image will be positive.

But a real image is formed below the principal axis, so the height (h_{2}) of a real image will be negative because it is measured in the downward direction. From this discussion we conclude that though the height of object h_{1} is always positive, the height h_{2} of the image can be either positive or negative.

Now, for a virtual image h_{2} is positive and h_{1} is also positive, so the magnification \(\left(\frac{h_2}{h_1}\right)\) of a virtual (and erect) image is always positive. In other words, if the magnification has a plus sign, then the image is virtual and erect.

For a real image, h_{2} is negative and h_{1} is positive, so the magnification \(\left(\frac{h_2}{h_1}\right)\) for a real (and inverted) image is always negative. Iri other words, if the magnification has a minus sign, then the image is real and inverted.

Since a concave mirror can produce virtual images as well as real images, the magnification produced by a concave mirror can be either positive or negative. A convex mirror, however, forms only virtual images, so the magnification produced by a convex mirror is always positive.

Another point to be noted is that if the magnification m has a value greater than 1 then the image is bigger than the object, that is, the image is magnified or enlarged. And if the magnification m is exactly 1, then the image is of the same size as the object. But if the magnification is less than 1 then the image is smaller than the object (or diminished).

A concave mirror can form images which are smaller than the object, equal to the object or bigger than the object, therefore, the linear magnification (or just magnification) (m) produced by a concave mirror can be less than 1, equal to 1 or more than 1.

On the other hand, a convex mirror forms images which are always smaller than the object, so the linear magnification (m) produced by a convex mirror is always less than 1. A plane mirror forms images which are always of the same size as the object, therefore, the magnification (m) produced by a plane mirror is always 1.

If we know the height (or size) of the object and that of the image, then we can calculate the magnification by using the formula given above. Many times, however, we do not know their heights, so we will now write another formula for calculating the magnification produced by a spherical mirror in terms of “object distance” and “image distance”.

The linear magnification produced by a mirror is equal to the ratio of the image distance to the object distance, with a minus sign. That is,

Magnification = – \(\frac{\text { Image distance }}{\text { Object distance }}\)

or m = –\(\frac{v}{u}\)

where m = magnification

v = image distance

and u = object distance

Thus, if we know the image distance v and the object distance u, then the magnification m can be calculated.

So, now we have two formulae for calculating the magnification :

m = \(\frac{h_2}{h_1}\) and m = – \(\frac{v}{u}\)

We will use these two formulae to solve numerical problems. We can also combine these two formulae to get another formula :

\(\frac{h_2}{h_1}\) = – \(\frac{v}{u}\)

This will also be used in solving problems.

### Numerical Problems Based on Concave Mirrors

We will now solve some numerical problems based on concave mirrors by using the mirror formula and the magnification formulae. Here are some examples.

**Example Problem 1.**

Find the size, nature and position of image formed when an object of size 1 cm is placed at a distance of 15 cm from a concave mirror of focal length 10 cm.**Solution:**

Here we have been given the object distance and focal length, so first of all we will find out the image distance which will give us the position of image.

(i) Position of image

Here, Object distance, u= -15 cm (To the left of mirror)

Image distance, v = ? (To be calculated)

And, Focal length, f = -10 cm (It is concave mirror)

Now, putting these values in the mirror formula :

\(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)

we get: \(\frac{1}{v}\) + \(\frac{1}{-15}\) = \(\frac{1}{-10}\)

or \(\frac{1}{v}\) – \(\frac{1}{15}\) = – \(\frac{1}{10}\)

or \(\frac{1}{v}\) = – \(\frac{1}{10}\) + \(\frac{1}{15}\)

\(\frac{1}{v}\) = \(\frac{-3+2}{30}\)

\(\frac{1}{v}\) = – \(\frac{1}{30}\)

So, Image distance, v = -30 cm

Thus, the position of image is 30 cm to the left side of mirror or 30 cm in front of mirror (Minus sign shows the left side of mirror).

(ii) Nature of image. Since the image is formed in front of the concave mirror, its nature will be “Real and Inverted”.

(iii) Size of image. To find the size of image, we will have to calculate the magnification first. The magnification produced by a mirror is given by :

m = – \(\frac{v}{u}\)

Here, Image distance, v = – 30 cm

Object distance, u = -15 cm

So, m = – \(\frac{(-30)}{(-15)}\)

m = – \(\frac{30}{15}\)

Magnification, m = -2

We also have another formula for magnification, which is :

m = \(\frac{h_2}{h_1}\)

Here, Magnification, m = -2 (Found above)

Height of image, h_{2} = ? (To be calculated)

Height of object, h_{1} = 1 cm (Given)

Now, putting these values in the above magnification formula, we get :

-2 = \(\frac{h_2}{1}\)

So, Height of image, h_{2} = – 2 × 1

= -2 cm

Thus, the size of image is 2 cm long. The minus sign here shows that the image is formed below the principal axis. That is, it is a real and inverted image.

**Example Problem 2.**

An object 2 cm high is placed at a distance of 16 cm from a concave mirror which produces a real image 3 cm high.

(i) What is the focal length of the mirror ?

(ii) Find the position of the image.**Solution:**

(i) Calculation of position of image

The height of object is 2 cm and that of real image is 3 cm. So, we will first calculate the magnification by using the formula :

m = \(\frac{h_2}{h_1}\)

Please note that an object is always placed above the axis, so the height of object is always taken as positive and written with a plus sign. A real image is formed below the axis. So, the height of a real image is taken as negative and written with a minus sign.

Here, Height of real image, h_{2} = – 3 cm

Height of object, h_{1} = + 2 cm

So, putting these values in the above magnification formula, we get:

Magnification, m = \(\frac{-3}{+2}\)

or m = -1.5

Now, we have another magnification formula for mirrors, which is :

m = – \(\frac{v}{u}\)

Magnification, m = – 1.5 (Found above)

Image distance, v = ? (To be calculated)

and Object distance, u = – 16 cm (To the left of mirror)

So, putting these values in the above formula, we get:

-1.5 = –\(\frac{v}{(-16)}\)

or -1.5 = \(\frac{v}{16}\)

So, v = -1.5 × 16

or v = -24 cm

Thus, the position of image is 24 cm in front of the mirror, to the left side of mirror (The minus sign shows that the image is on the left side of the mirror).

(ii) Calculation of focal length

Here, Object distance, u = – 16 cm

Image distance, v = – 24 cm

Focal length, f = ? (To be calculated)

Now, putting these values in the mirror formula :

\(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)

we get : \(\frac{1}{-24}\) + \(\frac{1}{-16}\) = \(\frac{1}{f}\)

or – \(\frac{1}{24}\) – \(\frac{1}{16}\) = \(\frac{1}{f}\)

\(\frac{-2-3}{48}\) = \(\frac{1}{f}\)

– \(\frac{5}{48}\) = \(\frac{1}{f}\)

f = – \(\frac{48}{5}\)

So, Focal length, f = – 9.6 cm

**Example Problem 3.**

A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located ?**Solution:**

In this problem we have been given magnification (m) and object distance (w). We are to find out the image distance (v). Now,

Magnification, m = – 3 (Image is real)

Object distance, u = -10 cm (To the left of mirror)

And, Image distance, v = ? (To be calculated)

Putting these values in the magnification formula for a mirror :

m = –\(\frac{v}{u}\)

we get: -3 = – \(\frac{v}{-10}\)

-3 × -10 = -v

v = – 30 cm

Thus, the image is located at a distance of 30 cm in front of the mirror (on its left side).

**Example Problem 4.**

The magnification produced by a plane mirror is +1. What does this mean ?**Answer:**

The plus sign (+) of the magnification shows that the image is virtual and erect. And the value 1 for magnification shows that the image is exactly of the same size as the object. So, the magnification of +1 produced by a plane mirror means that the image formed in a plane mirror is virtual and erect, and of the same size as the object.

**Example Problem 5.**

What is the nature of the image formed by a concave mirror if the magnification produced by the mirror is +3 ?**Answer:**

If the magnification has a plus sign (+), then the image is virtual and erect. In this case, the magnification has a plus sign (it is +3), therefore, the nature of this image is virtual and erect.

**Example Problem 6.**

What is the nature of the image formed by a concave mirror if the magnification produced by the mirror is, – 0.75 ?**Answer:**

If the magnification has a minus sign (-), then the image is real and inverted. In this case, the magnification has a minus sign (it is, – 0.75), so the nature of image is real and inverted.