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# Mirror Formula

Mirror Formula :- Below figure shows the ray diagram of image formation by concave mirror considering three rays. It shows the image A′B′ (real in this case) of an object AB.

In the above figure only three rays emanating from the point A are shown but actually an infinite number of rays emanate from any source, in all directions. Here, point A′ is image point of A if every ray originating at point A and falling on the concave mirror after reflection passes through the point A′.

Let us now derive the Mirror Formula or the relation between the object distance (u), image distance (v) and the focal length (f ).

In the figure the two right-angled triangles A′B′F and MPF are similar.(For paraxial rays, MP can be considered to be a straight line perpendicular to CP.) Therefore,

$\displaystyle \frac{B'A'}{PM}=\frac{B'F}{FP}$

or $\displaystyle \frac{B'A'}{BA}=\frac{B'F}{FP}(\because PM=AB)$ …..(1)

Since ∠ APB = ∠ A′PB′, the right angled triangles A′B′P and ABP are also similar. Therefore,

$\displaystyle \frac{B'A'}{BA}=\frac{B'P}{BP}$ …..(2)

From equation (1) and (2),

$\displaystyle \frac{B'F}{FP}=\frac{B'P-FP}{FP}=\frac{B'P}{BP}$ …..(3)

Now applying sign convention, we get….

B′P = –v,   FP = –f,   BP = –u

Using this sign convention in equation (3),

$\displaystyle \frac{-v+f}{-f}=\frac{-v}{-u}$

or

$\displaystyle \frac{v-f}{f}=\frac{v}{u}$

$\displaystyle \Rightarrow \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ …..(4)

This relation is known as the Mirror Formula.

Linear Magnification (m) :-

The ratio of the height of the image (h′) to the height of the object (h), is called linear magnification.

$\displaystyle m=\frac{h'}{h}$

h and h′ will be taken positive or negative in accordance with the sign convention. In triangles A′B′P and ABP, we have,

$\displaystyle \frac{B'A'}{BA}=\frac{B'P}{BP}$

Applying sign convention to the above equation, we get…

$\displaystyle \frac{-h'}{h}=\frac{-v}{-u}$

$\displaystyle m=\frac{h'}{h}=-\frac{v}{u}$ …..(5)

We have derived here the mirror formula, Eq. (4), and the linear magnification formula, Eq. (5), for the case of real, inverted image formed by a concave mirror.

With the proper use of sign convention, these are also valid for all the cases of reflection by a spherical mirror (concave or convex) whether the image formed is real or virtual.

Other forms of Linear Magnification:-

(1). In terms of v and f :-

from mirror formula,

$\displaystyle \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

multiplying above equation by v, we get

$\displaystyle \frac{v}{v}+\frac{v}{u}=\frac{v}{f}$

$\displaystyle 1+\frac{v}{u}=\frac{v}{f}$

$\displaystyle \Rightarrow \frac{v}{u}=\frac{v}{f}-1$

Hence, $\displaystyle m=-\frac{v}{u}=-\left( \frac{v-f}{f} \right)$

or

$\displaystyle m=\left( \frac{f-v}{f} \right)$ …..(6)

(2). In terms of u and f :-

from mirror formula,

$\displaystyle \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

multiplying above equation by u, we get

$\displaystyle \frac{u}{v}+\frac{u}{u}=\frac{u}{f}$

$\displaystyle \frac{u}{v}+1=\frac{u}{f}$

$\displaystyle \frac{u}{v}=\frac{u}{f}-1=\frac{u-f}{f}$

$\displaystyle \Rightarrow \frac{v}{u}=\frac{f}{u-f}$

Hence,

$\displaystyle m=-\frac{v}{u}=-\left( \frac{f}{u-f} \right)$

or

$\displaystyle m=\left( \frac{f}{f-u} \right)$ …..(6)

Hence, finally

$\displaystyle m=\frac{h'}{h}=-\frac{v}{u}=\frac{f}{f-u}=\frac{f-v}{f}$ …..(7)

Note :-

1. If m > 1, then size of image > size of object
2. If m < 1, then size of image < size of object
3. If m = 1, then size of image = size of object
4. If m = +ve  ⇒ then image is virtual and erect
5. If m = -ve  ⇒ then image is real and inverted

Illustration 1.

An object is placed at (i) 10 cm, (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature, and magnification of the image in each case.

Solution.

(a) In concave mirror, using sign convention

Using mirror’s relation

$\displaystyle \frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

$\displaystyle \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{2}{15}+\frac{1}{10}=\frac{-1}{30}$

v = -30 cm

Image is real and it forms at a distance of 30 cm in front of mirror.

Magnification

$\displaystyle m=-\frac{v}{u}=-\frac{(-30)}{(-10)}=-3$

This shows image is inverted and 3 times magnified.

(b) again using sign convention

$\displaystyle \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=-\frac{2}{15}+\frac{1}{5}=\frac{-1}{30}=\frac{1}{15}$

v = +15 cm

Image is formed at a distance of 15 cm behind the mirror.

Magnification

$\displaystyle m=-\frac{v}{u}=-\frac{(15)}{(-5)}=3$

+3 magnification shows image is virtual and erect.

Illustration 2.

Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of R = 2 m. If the jogger is running at a speed of 5 ms–1, how fast the image of the jogger appear to move when the jogger is (a) 39 m, (b) 29 m, (c) 19 m, and (d) 9 m away.

Solution.

From the mirror equation, we get

$\displaystyle v=\frac{uf}{u-f}$

For convex mirror, since R = 2 m, f = 1 m. Then

For u = -39 m, $\displaystyle v=\frac{uf}{u-f}=\frac{(-39)(1)}{(-39)-(1)}=\frac{39}{40}m$

Since the jogger moves at a constant speed of 5 ms–1, after 1 s the
position of the jogger u = –39 + 5 = –34m

So, For u = -34 m, $\displaystyle v'=\frac{uf}{u-f}=\frac{(-34)(1)}{(-34)-(1)}=\frac{34}{35}m$

The shift in the position of image in 1 s

x = v – v’ = (39/40) – (34/35) = 1/280 m

This shift occurs in 1 second. Hence, the average speed of the image when the jogger is between 39 m and 34 m from the mirror, is (1/280) ms–1.

Similarly, it can be seen that for u = –29 m, –19 m and –9 m, the speed with which the image appears to move is

(1/150)ms-1(1/60)ms-1(1/10)ms-1, respectively.

Another method

From the mirror equation $\displaystyle v=\frac{uf}{u-f}$

Differentiating above equation w.r.t. time, we get

$\displaystyle \frac{dv}{dt}=\frac{(u-f)\left[ f\frac{du}{dt} \right]-fu\left[ \frac{du}{dt} \right]}{{{(u-f)}^{2}}}$

$\displaystyle \Rightarrow \frac{dv}{dt}=-\frac{{{f}^{2}}}{{{(u-f)}^{2}}}\frac{du}{dt}$

(A)When jogger is 39m away,

$\displaystyle \frac{dv}{dt}=-\frac{{{1}^{2}}}{{{(-39-1)}^{2}}}(5)=-3.125\times {{10}^{-3}}m/s$

(B)When jogger is 29m away,

$\displaystyle \frac{dv}{dt}=-\frac{{{1}^{2}}}{{{(-29-1)}^{2}}}(5)=-5.56\times {{10}^{-3}}m/s$

(C)When jogger is 19m away,

$\displaystyle \frac{dv}{dt}=-\frac{{{1}^{2}}}{{{(-19-1)}^{2}}}(5)=-0.0125m/s$

(D) when 9m away,

$\displaystyle \frac{dv}{dt}=-\frac{{{1}^{2}}}{{{(-9-1)}^{2}}}(5)=-0.05m/s$

Note that negative sign implies opposite direction so if the object moves towards the mirror, it would have the image moving in the opposite direction (which would also be towards the mirror).

Next Topic :- Refraction of light

### About the author

#### Manoj Kumar Verma

Hi, I'm Manoj Kumar Verma, a physics faculty having 7 years of teaching experience. I have done B.Tech (E.E.). I am also a YouTuber and Blogger. This blog is dedicated to help students learn the physics concepts easily.

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