Magnification in spherical mirror, Practice problems, FAQs

As we know a plane mirror forms the image of the same size as an object but it is not true in the case of a spherical mirror. In a spherical mirror the size of the image can be smaller, bigger or the same as the object. To calculate the size of image we define a term called magnification which is the ratio of size of image to size of object. Now if the value of magnification is greater than 1 then size will be bigger, if less than 1, the size will be smaller and if it is 1, the size is equal. Lets see the various types of magnification and formula to calculate them!

Table of content

• Lateral or transverse magnification
• Longitudinal Magnification
• Areal Magnification
• Practice problems
• FAQs

Lateral or transverse magnification

For linear objects, lateral or transverse magnification is defined as the ratio of the height of the image to the height of the object. The heights of the object and the image are measured perpendicular to the principal axis.

Let us consider a concave mirror when an object of height ho is placed at a distance u from the mirror, and it produces an image of height hi at a distance v in front of the mirror as shown in the ray diagram.

Magnification is defined as,

Now let's find the relation between f, u and v.

Relation of Magnification in Terms of u and v

Let us consider a concave mirror and an object AB of height ho placed at a distance u from pole P, and it produces an image of height hi at a distance v in front of the mirror as shown in the ray diagram.

From the ray diagram, we get,

From and

And

So, from triangle and , we get, 

Hence, Magnification

Now we know from mirror formula

and

From equation Magnification

This is the formula for lateral magnification.

Magnification gives us many other important properties of the image formed by the mirror.

• For positive magnification, both the height of the object and the height of the image are of the same sign, i.e., both are positive or negative. If both ho and hi are positive, it means that the object is above the principal axis. This is possible only if the image is virtual and erect.
• For negative magnification, hi and ho are of different signs, If ho is positive then hi is negative and if hi is negative then we can conclude that the image is formed below the principal axis. So, the image formed is real and inverted.
• For positive magnification, u and v must be of the opposite signs, i.e., if u is positive, then v is negative and vice versa. Similarly, for negative magnification, u and v must be of the same sign, i.e., both u and v can be either positive or negative. For a real object, if the image formed is virtual, erect, and twice the height of the real object, then m = +2. Similarly, for a real object, if the image formed is real, inverted, and one-fourth the height of the real object, m =  

Longitudinal Magnification

For an object placed with its length along the principal axis, longitudinal magnification is defined as the ratio of the length of the image to the length of the object.

Mathematically, 

Longitudinal Magnification

Or

Longitudinal magnification for a small object

Let us assume that a small object of length du is placed in front of a concave mirror. Its image dv is formed in front of the mirror.

Longitudinal Magnification

From the mirror formula,

On differentiating with respect to u,we get,

is constant, so

For small object

This formula can be used for finding longitudinal magnification for small objects.

Areal Magnification

Let us consider a rectangular object of dimensions length a and breadth b placed in front of a concave mirror such that the length and breadth are perpendicular to the principal axis. The image is formed with magnification, the length changes to ma, and the breadth changes to mb, where m is lateral magnification.

Areal magnification is defined as the ratio of the area of the image (Ai) and the area of the object (Ao).

Areal Magnification

Area of the image, Ai = ma × mb = m2ab

Area of the object, Ao = ab

Magnification

This is formula for areal magnification

Practice problems

Q1. Find the magnification (m) and the position of the image in the figure if PO = 10 cm and the focal length of the mirror is 20 cm.

A. m = -2 and the image is 10 cm in front of the mirror.
B. m = -2 and the image is 60 cm behind the mirror.
C. m = 2 and the image is 20 cm in front of the mirror.
D. m = 2 and the image is 10 cm behind the mirror.

Answer. Given, In this case, the object is virtual.

Focal length, f = 20 cm

Object distance, u = 10 cm

We can get the position of the image by using,

Magnification

The image is formed 20 cm in front of the mirror, and m = 2

Thus, option (C) is the correct answer.

Q2. If the focal length of a concave mirror is 20 cm, find the position of a real object for which the real image is three times the size of the object. 

ANSWER. Given,

Focal length, f = -20 cm

It is given that the object and the image are real, so the magnification will be negative.

Therefore, m = -3

Substituting m in the magnification formula, we get the following:

The image is formed towards the left of the mirror.

Q3. If the focal length of a convex mirror is 20 cm, find the position of a real object for which the virtual image is times the size of the object. 

A. 120 cm toward the left of the mirror
B. 120 cm towards the right of the mirror
C. 80 cm towards the left of the mirror
D. 80 cm towards the right of the mirror

Answer. Given,

Focal length, f = 20 cm

It is given that the object is real and the image is virtual, so the magnification will be positive.

Therefore, m =  

Substituting m in the magnification formula, we get the following:

The image is formed 80 cm towards the left of the mirror.

Thus, option (C) is the correct answer.

Q4. An object of size 7.5 cm is placed in front of a concave mirror of radius of curvature 25 cm at a distance of 40 cm. Find the size of the image.

A. 3.41cm
B. 1.78cm
C. 1cm
D. 0.8cm

Answer.Given,

Radius of curvature, R = -25 cm

Object distance, u = -40 cm

Height of object, ho = 7.5 cm

Focal length,

Magnification,

The size of the image is -3.41 cm.

Thus, option (A) is the correct answer.

FAQs

Q1. What is the unit of magnification?
Answer: Magnification is the ratio of distance, so unit less quantity.

Q2. What is the magnification of an object placed at the center of curvature of a concave mirror
Answer: Magnification is -1 when the object is placed at the center of curvature.

Q3. The value of magnification can be less than 1?
Answer: Yes, for convex mirror images are always diminished, so magnification is less than 1.

Q4. How does focal length affect magnification?
Answer: Longer the focal length, angle of view will be narrow and magnification will be higher.