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The magnification produced by a spherical mirror gives a relative extent to which the image formed by an object is magnified with respect to the size of the object. A negative sign in the value of the magnification denotes that the image is real, while a positive sign in the value of the magnification denotes that the image is virtual. Let’s explore the mirror formula (1/f = 1/v +1/u) and analyse the way to locate images without drawing any ray diagrams.
We have often used a pair of binoculars to observe faraway objects. While observing through the device, we witness that the size of the objects seen through the binoculars are not the same as we see them with our naked eyes. This is because the binoculars focus the light from distant objects and help us to view the object in greater detail by making them appear bigger. This was an example of mirror formula and magnification. ?
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Keyterms: Magnification, Mirror, Lens, Converging mirror, Object, Image, Spherical Mirror, Focal Length, Radius
Definition
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Mirror Formula
The Mirror formula represents the relationship between the distance of the object (u), the distance of the image (v), and the focal length (f) of a Spherical Mirror. Mirror formula is written as:
1/f = 1/v+1/u
Here, u and v are the Distance of the Object and image from the pole of the Mirror Respectively. And Focal Length (f) is the principal focus Distance from the pole.
We can determine the magnification(m) of the object with the help of u and v by using the equation below.
m=−v/u
The radius of curvature (R) is twice its focal length.
R=2f
and f=R/2
Hence, the Mirror Formula can be written as:
1/u+1/v = 1/f= 2/R
The video below explains this:
Mirror Formula Detailed Video Explanation:
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Explanation
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After falling on a highly polished surface like a mirror, light gets reflected. The spherical mirror reflecting surface can be curved inwards or outwards. A spherical mirror that has a reflecting surface curved inwards is known as a concave mirror. A spherical mirror that has a reflecting surface curved outwards is known as a convex mirror.
Concave mirrors are also called converging mirrors because of the rays getting converged after falling on the concave mirror while the convex mirrors are called diverging mirrors because of the rays getting diverged after falling on the convex mirror.
In a spherical mirror, the distance of the object and image from its pole is called the object distance(u) and the image distance (v) respectively. We also know that the focal length(f) is the distance of the principal focus from the pole.
1/f = 1/v+1/u
This formula is accurate in all situations for planes as well as for all spherical mirrors for all positions of the object.
Sign Convention for Spherical Mirrors
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Sign Convention for Spherical Mirrors
The sign convention for spherical mirrors has a set of rules to be followed that is known as the “New Cartesian Sign Convention" as described below:
- The pole(p) of the mirror acts as the origin.
- The principal axis acts as the x-axis of the coordinate system.
- The object is kept on the left of the mirror. This is to be followed in all situations.
- The distances that are parallel to the principal axis are to be measured from the pole(p).
- All the distances measured from the pole (p) on the right-hand side of the mirror are taken to be positive and those on the left-hand side of the mirror are taken to be negative.
- Distances that are perpendicular and lying above the principal axis are to be taken positive.
- All the distances below the principal axis are to be taken negative.
Magnification
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The size of the image formed by a mirror is called Magnification. We can also find whether the image formed will be magnified, diminished, or equal to the object.
Magnification is represented by the ratio between the height of the image to that of the object. It is denoted by “m”. It doesn't have any unit and can be expressed as:
m=h′/h
Here, h′ is the height of the image and h is the height of the object.
Magnification is also equal to the negative of the ratio of the image distance(v) to object distance(u).
m=−v/u
Here, height of the object is positive as the object is usually above the principal axis. But a sign for height of the image varies according to the type of image produced.
To have a better understanding of the topic, refer to this video:
Things to Remember
- The mirror formula is 1/f = 1/v + 1/u
- The magnification m is denoted by m = -vu.
- According to the sign conventions, the focal length (f) for a concave mirror is negative, the object distance (u) is negative.
- According to the sign conventions, the image distance (v) is positive for a virtual image while negative for a real image.
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Sample Questions
Ques: What is the magnification produced if the image distance is 6 cm and the object is located at 12 cm in case of concave mirror? (2 marks)
Ans: We already know the magnification can be calculated using the following formula: m = -v/u
Given, v = -6 cm and u = -12 cm
The signs are given byy using the sign convention.
therefore, m = – (-6cm/ -12cm)
or, m = -1/2
or, m = -0.5
Thus, there will be a decrease by a factor of 0.5.
Ques: Find the image distance in case of convex mirror if the object is placed at 12 cm. Determine it if the height of the image is 4 cm and height of the object is 2 cm. (3 marks)
Ans: As we know the magnification can be calculated using the following formulas: m = -v/u and also m = h’/h
Given, the height of the image h’ = 4 cm, height of the object (h) = 2 cm and u = -12 cm
The signs are given using sign convention.
m = h’/h
or, m = 4 cm/ 2 cm
m = +2
Thus, there is an increase by a factor of 2.
m = -v/u
Putting m = 2 and u = -12 cm we get,
2 = – (v/(- 12 cm)
or, v = (-2) x (- 12 cm)
or, v = 24 cm
Thus, the image distance is 24 cm.
Ques: What is the magnification if the object height is 6 cm and the image height is 24 cm below the principar axis? (2 marks)
Ans: As we know the magnification can be calculated using the following formulas: m = h’/h
Given, height of the image h’ = -24 cm, height of the object (h) = 6 cm
The signs are given using sign convention.
therefore, m = h’/h
or, m = -24 cm/ 6 cm
0r, m = -4
Thus the magnification is -4.
Ques: What is the image distance of concave mirror if the object distance is 8 cm? It is given that the focal length of the mirror is 4 cm. (3 marks)
Ans: As we know the mirror formula: 1/v + 1/u = 1/f
where, u = object distance = -8 cm
v = image distance = ?
f = focal length of the mirror = -4 cm
Putting the values we get,
1/v + 1/-8 = 1/-4
or, 1/v = -1/8
or, v = -8 cm
Therefore, the object is located 8 cm in the front mirror.
Ques: Find the image distance of convex mirror if the object distance is 10 cm? It is given that the focal length of the mirror is 10 cm. (3 marks)
Ans: As we know the mirror formula: 1/v + 1/u = 1/f
where, u = object distance = -10 cm
v = image distance = ?
f = focal length of the mirror = =10 cm
Putting the values we get,
1/v + 1/-10 = 1/10
or, 1/v = 1/10 – (1/-10)
or, 1/v = 1/10 + 1/10
or, 1/v = 2/10
or, 1/v = 1/5
or, v = 5 cm
Therefore, the image is located 5 cm behind the mirror.
Ques: Amit kept a pencil perpendicular to the principal axis of a concave mirror just ahead of it. The Mirror has a focal length of 30 cm. The image produced is two times the pencil size. Assess the object distance from the mirror. (5 marks)
Ans: Magnification => h(i)/h(o) =−v/u
For real image
m=−v/u=−2
v=2u
Using the mirror equation,
1/v+1/u=1/f
1/2u+1/u=1/−30
u=-45 cms.
This lies between the focal length and the Curvature.
For virtual image
m =−v/u=2
v=-2u
Using the mirror equation,
1/v+1/u=1f
(1/−2u)+(1/u)=1/-30
u=-15 cm.
Ques: An Object is kept at a distance of 25cm in front of a concave mirror which has a focal length of 15 cm. At what distance from the Mirror, we should place a screen to obtain a sharp image? (2 marks)
Ans: Given. f=-15cm, v= -30 cm
Using Mirror Formula,
1/v+1/u=1/f
1/u=1/f-1/v
1/u=(1/-15) - (1/-30)
1/u = -1/30
u=-30 cm.
Ques: Suppose that Height of an Object is 6 cm. Height of the image is 24 cm. What is magnification? (2 marks)
Ans: Given, Height of image is (h')=24 Cm
Height of object (h) = 6 cm
So, m = h'/h
= 24/6
= 4
Hence, Magnification is 4.
Ques: A convex Mirror Used to see rear-view on a Car has a radius of Curvature of 3m. Another Car is located 5m from this Mirror. Assess the image. (5 marks)
Ans: Radius of Curvature = 3m
We know that, f = R/2
=3/2
=1.5m
The Focal Length will be Positive as the focus of Convex Mirror is behind the Mirror.
Another car (o) Distance will be negative,
U=-5m
Let the image Distance be v
Using Mirror Formula,
1/f=1/v+1/u
1/f-1/u=1/v
1/v=1/f-1/u
1/v=1/1.5+1/5
1/v=10/15+1/5
1/v= 10+3/15
1/v=15/13
V=+1.15m
The Image is formed at a Distance of 1.15m and Behind the Mirror.
For Size of Image, m=-v/u
= -1.15/-5
= 115/500
= +0.23
Magnification is less than 1, so Image is diminished.
Image formed is erect as Magnification is Positive.
So, we conclude that the image is Virtual, Erect and diminished in Size.
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