Contents

**Magnification**

### Introduction

An image is said to be **magnified **when its size is **greater** than that of the object and it is said to be **diminished** when its size is **smaller** than that of the object. Magnification is a physical quantity used to measure the amount of magnification of an image as compared to an object. In this session we will be looking at the concept of magnification and solve numerical problems based on it.

### Explanation

**Magnification:**

Magnification is a measure of how large or small in size the image of an object is, compared to that of the object.

The magnification ‘m’ is mathematically given by,

In the above formula, the height of the object is taken to be positive, as the object is usually placed above the principal axis. The height of the image is taken to be negative for a real image and positive for a virtual image.

The magnification is a **unitless** physical quantity.

**Magnification **is also related to the object distance (**u**) and the image distance (**v**) as follows.

**Case 1: For a magnified image: **

For a magnified image, the size of the image is greater than the size of the object.

That is H > h.

,Therefore,

gives, **m >1.**

**Case 2: For a diminished image: **

For a diminished image, the size of the image is smaller than the size of the object.

That is H < h.

,Therefore,

gives, **m <1.**

**Case 3: For a same sized image: **

For an image of the same size as the object,

H = h.

Therefore,

gives, **m =1.**

**Numerical Problems:**

**An object of height 6 cm is placed at a distance of 5 cm in front of a concave mirror of focal length 2.5 cm. How magnified/diminished will the image of this object be?**

The height of the object = 5 cm

Object distance = 5 cm

Focal length = 2.5 cm.

The radius of curvature = 5 cm

Therefore, the object is located at the radius of curvature of the mirror.

Thus, a real, inverted and same sized image is formed at the center of curvature itself.

Therefore, the size of the image = – 6 cm.

Magnification,

i.e., m =

Therefore, **magnification is – 1**. The negative sign indicates the real and inverted nature of the image.

**An object of height 10 cm is placed at a distance of 30 cm from a mirror of focal length 25 cm. The image of this object is found to be of 15 cm height. What is the position of the image?**

Object height = 10 cm

Image height = – 15 cm (inverted)

Object distance = 30 cm

Image distance = ?

Magnification = – 15/10 = **-1.5**

The image is formed at a distance of 45 cm in front of the mirror and is magnified 1.5 times that of the object.

**An object 5 cm in size, is placed at 25 cm in front of a concave mirror of focal length 15 cm. at what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and the size of the image.**

Object size = 5 cm

Object distance = – 25 cm

Focal length = – 15 cm

Image distance = ?

Size of the image = ?

Nature of the image = ?

From the mirror formula we have,

The object is located in front of the mirror at a distance of 37.5 cm. Thus, the screen needs to be placed at a distance of **37.5 cm **in front of the mirror to get a sharp image.

Thus, the size of the image is **7.5 cm**.

A real, inverted and magnified image of size **7.5 cm **is formed at a distance of **37.5 cm **in front of the mirror.

**Summary**

- Magnification is a measure of how large or small in size the image of an object is,compared to that of the object.
- The magnification ‘m’ is mathematically given by,
**m = height of the image / height of the object**i.e - The magnification is a
**unitless**physical quantity. - The magnification, m < 1, for a diminished image, m > 1, for a magnified image andm= 1 for a same sized image.
**Magnification**is also related to the object distance**(u)**and the image distance**(v)**as,- A
**negative**value of magnification indicates a**real and inverted**image. - A
**positive**value of magnification indicates a**virtual and upright**image.