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Mirror formula, also called the mirror equation, is an equation which relates object distance and image distance with focal length. The Mirror formula is used to obtain the numerical information about image distance and image size that ray diagrams fail to provide. The Mirror formula also establishes a relationship between the object distance (v), image distance (u) and focal length (f).
| \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\) |
- f = focal length of the mirror
- u = distance of the object from the mirror (measured along the principal axis and on the object side of the mirror)
- v = distance of the image from the mirror (measured along the principal axis and on the image side of the mirror)
The mirror formula helps to calculate the position and size of the image formed by a spherical mirror for a given object distance and focal length.
| Table of Content |
Key Terms: Mirror Formula, Magnification, Derivation Of Mirror Formula, Sign Convention, Concave Mirror, Convex Mirror, Focal length, Principal Axis, Image Distance, Object Distance
Mirror Formula and Magnification Equation
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The Mirror Formula expresses the quantitative relationship between the object distance (v), the image distance (u), and the focal length (f). The equation for mirror formula is:
| \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\) |
The video below explains this:
Mirror Formula Detailed Video Explanation:
Magnification Equation: It relates the ratio of the image distance and object distance to the ratio of the image height (hi) and the object height (ho). The magnification formula of mirror is:
| \(M = \frac{h_i}{h_0} = - \frac{d_i}{d_0}\) |
Together, these two equations can be combined to yield information about the image distance and image height if the object distance, object height and focal length are known.
However, focal length can also be calculated by the formula,
→ f = R/2
Where, R is the radius of curvature of the spherical mirror.
Mirror Equation
In a spherical mirror:
- Object Distance (u) = Distance between the object and the pole of the mirror.
- Image Distance (v) = Distance between the image and the pole of the mirror.
- Focal length (f) = Distance between the Principal focus and the pole of the mirror.
Concave Mirrors
For Concave Mirrors, the following can be seen:
| Object Position | Image Position | Size of Image | Natural of the Image |
|---|---|---|---|
| Within focus ( Between pole P and focus F) | Behind mirror | Enlarged | Virtual, Erect |
| At focus | At infinity | Highly Enlarge | Real, Inverted |
| Between F and C | Beyond C | Enlarged | Real, Inverted |
| At C | At C | Equal to object | Real, Inverted |
| Beyond C | Between F and C | Diminished | Real, Inverted |
| At Infinity | At focus | Highly diminished | Real, Inverted |
Convex Mirrors
And, for Convex Mirrors,
| Object Position | Image Position | Size of the Image | Image Nature |
|---|---|---|---|
| Anywhere between pole P and Infinity | Behind the mirror between P and F | Diminished | Virtual and erect |
| At infinity | Behind the mirror at Focus | Highly diminished | Virtual and erect |
Read More:
| Learn more about mirrors | ||
|---|---|---|
| Ray Optics and Optical Instruments | Mirror Formula Derivation | Uses of Convex Mirror |
| Uses of Concave Mirror | Total Internal Reflection | Refraction |
Cartesian Sign Convention
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Cartesian sign convention is a set of rules for sign conventions that are widely used in optics, especially in context of lenses and mirrors. The Sign convention is generally based on the Cartesian coordinate system. It is used to find the signs of object distance (u), image distance (v), and focal length (f) in the mirror and lens formulas. The convention consists of the following rules:
- The object distance, u, is positive in case the object is on the same side of the mirror as the incident light (meaning that the object is in front of the mirror or lens), and negative case the object is on the opposite side of the mirror (meaning that the object is behind the mirror or lens).
- The image distance, v, is positive in case image is on the opposite side of the mirror as the incident light (meaning that the image is in front of the mirror or lens), and negative in case image is on the same side of the mirror as the incident light (meaning that the image is behind the mirror or lens).
- The focal length, f, is positive for converging lenses and concave mirrors, and negative for diverging lenses and convex mirrors.
The optical center of the lens is considered for measurement of all the distances.
- The distances measured opposite to the direction of the incident light are considered to be negative, while the distances measured in the same direction of the incident light are considered to be positive.
- The heights measured upwards and perpendicular to the principal axis are considered to be positive, while the heights measured downwards and perpendicular to the principal axis are considered to be negative.
How to Apply Mirror Formula and Magnification Equation?
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To demonstrate the application of the mirror formula and magnification equation, given below is a sample problem with the solution and procedure explained.
ProblemA 4.00-cm tall light bulb is placed a distance of 45.7 cm from a concave mirror having a focal length of 15.2 cm. Determine the image distance and the image size. SolutionStep 1- Begin by the identification of the known information. ho = 4.0 cm do = 45.7 cm f = 15.2 cm Step 2- Identify the unknown quantities that you wish to solve for. di =? hi =? Step 3- To determine the image distance, the mirror equation must be used. The following lines represent the solution to the image distance. Substitutions and algebraic steps are shown. 1/f = 1/do + 1/di 1/(15.2 cm) = 1/(45.7 cm) + 1/di 0.0658 cm-1 = 0.0219 cm-1 + 1/di 0.0439 cm-1 = 1/di di = 22.8 cm The final answer is rounded to the third significant digit. Step 4- To determine the image height, the magnification equation is needed. Since three of the four quantities in the equation (disregarding the M) are known, the fourth quantity can be calculated. The solution is shown below. hi/ho = - di/do hi / (4.0 cm) = - (22.8 cm)/ (45.7 cm) hi = - (4.0 cm) • (22.8 cm)/ (45.7 cm) hi = -1.99 cm |
Observations and Conclusions
- The negative values for image height indicate that the image is an inverted image.
- A negative or positive sign in front of the numerical value for a physical quantity represents information about direction.
- In the case of the image height, a negative value always indicates an inverted image.
- From the calculations in this problem it can be concluded that if a 4.00 cm tall object is placed 45.7 cm from a concave mirror having a focal length of 15.2 cm, then the image will be inverted, 1.99cm tall and located 22.8 cm from the mirror.
- The results of this calculation agree with the principles discussed earlier in this lesson.
- In this case, the object is located beyond the center of curvature (which would be two focal lengths from the mirror), and the image is located between the center of curvature and the focal point.
Uses of Magnification
Some of the magnification uses include:
- A precision magnifier serves as a simple magnifier, holding a variety of elements to erase the aberrations, thus forming a sharper image.
- A tiny water droplet acts as a very simple magnifier that magnifies the object present behind it. The water forms small spherical droplets due to the influence of the surface tension. When the water droplet is in contact with any object, a spherical shape is distorted but capable of forming a good image of the object.
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Things to Remember
- Mirror formula is used to establish a relationship between object distance, image distance and focal length.
- Mirror formula is \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\).
- Focal length is negative for concave lens and concave mirror.
- Image distance (v) can be both positive and negative for convex lens and concave mirror depending on the position of the object.
- Image distance is always negative for a concave lens. Image distance is always positive for a convex mirror.
- Convex mirror always forms an erect and virtual image. The size of the image is smaller as compared to the object.
- In a concave mirror, when the distance of the object is less than the focal length, the magnification will be greater than one. When the distance of the object is greater than the focal length, then the magnification is less than one.
Sample Questions
Ques. The image of a candle flame placed at a distance of 30 cm, space, start text, c, m, end text from a mirror is formed on a screen placed in front of the mirror at a distance of 60 cm, space, start text, c, m, end text from its pole. What is the nature of the mirror? Find its focal length. (3 marks) (Foreign 2010)
Ans. Since the image forms on a screen, it's a real image. The mirror is concave.
We are given:
object distance, u = -30
image distance, v = -60
The focal length, f, is calculated from the mirror formula,
f = -20 cm
Ques. A 6 cm object is placed perpendicular to the principal axis of a convex lens of focal length 15 cm. The distance of the object from the lens is 10 cm. Find the position, size and nature of the image formed, using the lens formula. (3 Marks)
Ans. Given that f=15cm , u= -10cm , v=?
Using Lens formula,
v = -30 cm
Hence the image is at a distance of 30 cm from the lens.The negative sign indicates it is on the same side of the lens as the object and it is a real image.
Now the size can be obtained using the magnification formula
hi = 18 cm
Hence, the position of the image is 30 cm on the same side of the lens and the image is 18 cm and it is an erect image.
Ques. What is the formula for spherical mirrors for object distance p and image distance q? (3 Marks)
a)1/p + q = 1/f
b)1/p + 1/q = 1/f
c)1/p + 1/q = f
d)p + 1/q = 1/f
Ans. Mirror formula is given as:
2/R = 1/v + 1/u
where, R is the radius of curvature of the spherical mirror
u is the object distance from the pole
v is the image distance from the pole
We know,
f = R/2
1/f = 1/v + 1/u
∴ For object distance p and image distance q, mirror formula becomes
1/f = 1/p + 1/q
Hence, the correct answer is OPTION B.
Ques. Why is the present sign convention known as the new sign convention? (4 Marks)
Ans. The sign convention followed for spherical mirrors and spherical lenses that is similar to sign convention of cartesian coordinate systems, called New Cartesian sign convention. It is easy to follow and easy to remember because we are very much familiar with the Cartesian coordinate system.
It is assumed in this sign convention, the pole of the mirror or optical centre of the lens is at the origin of the cartesian coordinate system. Distances parallel to the optical axis are measured by taking the pole of the mirror or optical centre of lens because origin is there at the pole of mirror or optical centre of lens.
Distances parallel to the optical axis which are right of the pole of mirror or optical centre of lens are +ve. Distances parallel to the optical axis which are left of the pole of mirror or optical centre of lens are -ve.
Distances perpendicular to optical axis are measured by taking optical axis as reference and given sign as per cartesian coordinate system, i.e., distances measured upward from optic axis are +ve and distances measured downward from optic axis are negative.
Ques. Calculate the image distance and magnification for a 5.00-cm tall object placed 10.0 cm from a concave mirror with a focal length of 5.0 cm. (5 Marks)
Ans. The focal length (f) = 5 cm
The object distance (do) = 10 cm
Formation of image by concave mirror :
The image distance:
1/di = 1/f – 1/do = 1/5 – 1/10 = 2/10 – 1/10 = 1/10
di = 10/1 = 10 cm
The image distance is 10 cm.
The magnification:
m = –di / do = -10/10 = -1
This means that the image is the same as the object.
The minus sign indicates that the image is inverted. If the sign is positive then the image is upright.
Ques. The focal length of a convex mirror is 10 cm and the object distance is 20 cm. Determine (a) the image distance (b) the magnification of image. (5 Marks)
Ans. The focal length (f) = -10 cm
The minus sign indicates that the focal point of the convex mirror is virtual.
The object distance (do) = 20 cm
Formation of image by concave mirror:
The image distance (di) :
1/di = 1/f – 1/do = -1/10 – 1/20 = -2/20 – 1/20 = -3/20
di = -20/3 = -6.7 cm
The minus sign indicates that the image is virtual.
The magnification of image :
m = – di / do = -(-6.7)/20 = 6.7/20 = 0.3
m = 0.3 time smaller than the object.
The plus sign indicates that the image is upward.
Ques. A point object is placed at a distance of 10cm and its real image is formed at a distance of 20cm from the concave mirror. If the object is moved by 0.1cm towards the mirror, the image will shift by about: (5 Marks)
0.4cm away from the mirror
0.4cm towards the mirror
0.8cm away from the mirror
0.8cm towards the mirror
Ans. We know 1/f = 1/u + 1/v;
where
f= focal
length = ?
u = initial distance = 20 cm
v = final distance = 10 cm
1/f = 1/-20 + 1/(-10)
f = 20/3 cm
Now, the object is moved towards the mirror = 0.1 cm, so the new value of u = 9.9 cm.
Again,
1/f = 1/u1 + 1/v1
1/v1 = 3/20 - 1/(-9.9)
v1 = 20.4 cm
The change in final distance = 20.4 cm - 20 cm = 0.4 cm
v = 9 x 1600/3600 = 4 cm/s
Ques. State the two laws of reflection of light. (2 marks) (Delhi 2011)
Ans. Laws of reflection of light states that
(i) The angle of incidence is equivalent to the angle of reflection.
(ii) The incident ray, the reflected ray and the normal to the mirror at the point of incidence can be seen to lie on the same plane.
Ques. Consider a image which is formed by a spherical mirror for all positions of the object in front of it is always erect and diminished. Assuming the same, determine the type of mirror it is. Also draw a labelled ray diagram to demonstrate your answer. (2 marks) (2018)
Ans. It is a convex mirror considering that the image formed by a spherical mirror is always erect and diminished. It can be demonstrated as:
Ques. Draw a ray diagram order to demonstrate the path of the reflected ray corresponding to an incident ray that has been directed parallel to the principal axis of a convex mirror. Here, make the angle of incidence and the angle of reflection. (2 marks) (Delhi 2014) (AI 2019)
Ans. The ray diagram can be shown as:
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