Solve Problems on Mirror and Magnification Formula

Binoculars are usually used to observe distant objects. While looking through this device, it has been noticed that the size of the objects visible via the binoculars differs from what we observe with our naked eyes. This is due to the fact that binoculars concentrate light from distant objects, allowing us to see them in more detail by making them look larger. This was a demonstration of the mirror formula and magnification.

Mirror Formula

The mirror formula shows the relationship between the object distance, image distance, and the focal length of the spherical mirror. Hence, the formula is given as,

1/u + 1/v  = 1/f

where, 

• u is the object distance, 
• v is the image distance, and 
• f is the focal length of the mirror

While solving numerical problems using mirror formula one must remember two important things according to the sign convention for spherical mirrors, that are:

1. If object is at left side of the mirror the object distance is taken as negative, if it is in right it is positive.
2. For focal length the sign is depend on type of mirror we are using if mirror is concave then focal length is taken negative and if mirror is convex then focal length is positive.

Mirror formula in terms of Radius of Curvature:

Since, the radius of curvature (R) is two times its focal length (f), that is

R = 2f

or 

f = R/2

Hence, the Mirror Formula can be written as:

1/u + 1/v = 1/f = 2/R

Magnification

One must have noticed that there is simply an increase or decrease in the size of image by concave or convex mirror, this is a magnification of the object. 

• Magnification is defined as the ratio of the height of the image (h_i) to the height of the object (h_o).

Magnification = Height of image / Height of object

or

m = H_i  / H_o

Magnification can also be defined as the ratio of the image distance (v) and the object distance from the mirror (v).

Magnification = -Image distance / Object distance 

or

m = -v / u

Determination of the Nature of the Image formed by a spherical mirror:

• The positive magnitude of magnification shows us that a virtual and erect image is formed.
• The negative  magnitude of magnification shows us that a real and inverted image is formed.

Problems on Mirror Formula and Magnification Formula

Problem 1: An object is placed at a distance of 2 times of focal length from the pole of the convex mirror, Calculate the linear magnification.

Solution:

Let the Focal length of mirror = f

So, the object distance, u = -2f 

The formula to calculate image distance we use mirror formula as,

1 / v + 1 / u = 1 / f

Therefore,

1 / v + 1 / -2f = 1 / f 

             1 / v  = 1 / f + 1 / 2f  

                      = 3 / 2f

or

v = 2f / 3

Magnification is given as, 

m = – v / u

= -(2f/3) / (-2f) 

=  1/3

Problem 2. If the image is a distance of 6 cm and the object is at 12 cm in the front of the concave mirror, Calculate the magnification formed.

Solution:

Given that,

The distance of object, u = – 12 cm

The distance of image, v = – 6 cm

Since,

Magnification is given by, 

m = – v / u

Therefore,

m = – (-6 / -12) 

= -0.5

Hence, the image will be diminished by nearly half as size of object. 

Problem 3: In the experiment height of the image is 12 cm whereas the height of the object is 3 cm, would you determine the magnification formed.

Solution:

Given that, 

Height of image = 12 cm 

Height of object = 3 cm

Magnification in terms of height is given by,

m = height of image / height of object

= 12 / 3 

= 4

Therefore magnification is 4.

Problem 4: In the case of a concave mirror if the object is placed at the distance of 12 cm. Determine the image distance from the mirror if the height of the object to image ratio is 1:2.

Solution:

Given that,

The object distance, u = -12 cm

Ratio of object to image height = 1/2

Magnification = height of image / height of object  

= 1/ (1/2) 

= 2

Now, magnification in terms of distance of object and image from the mirror,

m = – v / u

= – v / -12 

2 = v / 12  

or 

v = 12 × 2 

= 24

Therefore the distance of image from the mirror is equal to 24 cm.

Problem 5: Calculate the change in the size of the image formed, if the object distance is 18 cm and the distance of the image is 6 cm from the concave mirror. 

Solution:

Given that,

The object distance, u = -12 cm

Image distance, v = – 6cm

Magnification,  m = – v/u 

= – (-6 / -18) 

= -1/3

which means that size of image is 1/3^rd of the size of object.

Problem 6: The radius of curvature of the rear view convex mirror of the truck is 6 m. If the car is 8 m from the mirror of the truck. Calculate the distance at which the image is formed.

Solution:

Given that,

Radius of curvature, R = 6 m 

Object distance, u = -8 m

Focal length is half of Radius of curvature,

f = R/2 

=  6/2 

= 3 m

Using mirror formula 

1 / v + 1 / u = 1 / f

1 / v + 1 / -8 = 1 / 3

1 / v  = 1 / 3 + 1 / 8

= 11 / 24

v = 24 / 11 m

The image is formed at distance of 24 / 11 behind the mirror.

Problem 7: A concave mirror produces an image of size n times that of the object and of focal length f. If the image is real then find the distance of the object from the mirror.

Solution:

Given that

Size of image = n × size of object

n = Size of image /  size of object  = magnification

Since the image is real, it must be inverted hence magnification will be negative, 

m = -n

Let d is the distance of object then,

m = -v/u  

-n = -v / d 

or

v = nd

Therefore, the mirror formula:

1 / f = 1/v + 1/u

becomes,

1/f = 1/nd + 1/d

or

1/f = 1/d(1/n + 1)

or

1/d = n/ f(n + 1)

Therefore,

d = f (n + 1)/ n

Problem 8: Where should the object be placed to obtain a magnification of 1/3? If an object is placed at a distance of 60 cm from a convex mirror, then the magnification produced is 1/2. 

Solution:

Given that,

u = -60 cm

m  = 1/2 

So,

-v/u = 1/2 

and 

v/60 = 1/2

or

v = 30 cm

Since, the mirror formula is:

1 / v + 1 / u = 1 / f

Therefore,

1 / 30 + 1 / (-60) = 1/f

1/f = ( 2-1 ) / 60 = 1 / 60

f = 60 cm

Now for magnification = 1 / 3,

– v / u = 1 / 3

or 

v = – u / 3

 using mirror formula

 1 / v + 1 / u = 1 / f

1 / (-u/3) + 1/ u = 1/ 60

-3/ u + 1/u = 1/60

-2/ u = 1/60

or 

u = -120 cm

object should be placed at 120 cm in front of mirror to get magnification of 1/3.

Problem 9: In the case of a concave mirror, if the object distance is 11 cm, its focal length is 11 cm then, Calculate the image distance.

Solution:

Given that,

Distance of object, u = -11 cm

Focal length, f = -11cm

Using mirror formula,

1 / v + 1 / u = 1 / f

Therefore,

1 / v + 1 / -11 = 1/ -11

So,

1/v = 0

or 

v = infinity

This means that image will be at infinity if object is present at the focal length.

Problem 10: If the object distance is 32 cm in front of the concave mirror, the focal length of the mirror is 16 cm. State the nature and the size of the image formed.

Solution:

Given that,

Object distance, u = -32 cm

Focal length , f = -16 cm

For image distance use mirror formula,

1 / v + 1 / u = 1 / f

Therefore,

1/ v + 1/ -32 = 1/ -16

or 

1/ v = 1/ -16 + 1/ 32

or

1/ v = -1 / 16

So,

v = -16 cm

Hence the image is located 16 cm in front of the mirror. and the image formed is real and inverted. 

Size of image will be same as that of object, as it is located at center of curvature.