Question

Derive $\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$ formula for a spherical mirror.

Answer 1

Hint
The formula which we have been told to derive is known as the mirror formula. We all know that when an object is placed in front of a mirror, the image is formed behind the mirror as far as the object is in front of it. But that analogy is applicable only for plane mirrors. The reflecting property of spherical mirrors is different, hence we need the formula to calculate the image distance for any given object.

Complete step by step answer
For ease of communication, we will refer to spherical mirrors simply as mirrors from now on. We will derive this formula using a concave mirror which is curved inwards, but the same method is applicable for convex mirrors and all spherical mirrors in general.
The terms used in the formula are the image distance from the mirror, the object distance from the mirror and the focal length of the mirror. The figure below shows the image formation by a spherical concave mirror. We’ll use the figure to proceed with our derivation of the expression.

The parallel ray of light passes through the principal focus F after reflection and the ray of light appearing to diverge from the centre of curvature C suffers normal reflection.
The object is denoted by AB and the image is denoted by A’B’. For sake of your convenience, the object and the image are also labelled in the diagram.
Let the distance of the object from the mirror be $u$ , that is $u=PB$
The image distance from the mirror is given as $v$ where $v=-B'P$
The focal length of the mirror is given by $f=-PF$
The radius of curvature of the mirror is given as $R=-PC$
The sign convention is taken into account while writing the distances, the distances measured along the light ray are taken as positive whereas the distances measured opposite to the light rays are taken as negative.
Now, if we look at the triangles formed by the points ABC and A’B’C, we can use the property of vertically opposite angles to say that $\angle ACB=\angle A'CB'$
Now, in both the triangles we have one right angle, so we can say that $\angle ABC=\angle A'B'C$
If the two angles of a pair of triangles are equal, the third angles of both triangles will automatically be equal, that is $\angle BAC=\angle B'A'C$ . Hence, by AA or AAA law of triangle similarity, we can say that $\angle ACB\cong \angle A'CB'$
Now, we can use the property of similar triangles, which states that the ratio of the corresponding sides in a similar triangle must be equal, to say that
$\dfrac{AB}{A'B'}=\dfrac{BC}{B'C}-(1)$
We can use the same approach to state the triangles FED and A’B’F are similar to each other since both the triangles have one pair of vertically opposite angles and one right angle each.
Using the property of similar triangles, we can now say that
$\dfrac{ED}{A'B'}=\dfrac{EF}{FB'}-(2)$
From the expressions marked one and two, we can say that
$\dfrac{BC}{B'C}=\dfrac{EF}{FB'}$
Now, the points D and P are very close to each other since the aperture of the mirror is very small. So we can say that $EF=PF$
Now our derived expression would look something like $\dfrac{BC}{B'C}=\dfrac{PF}{FB'}$
In the above expression, we can put $BC=PC-PB$ , $B'C=PB'-PC$ and $FB'=PB'-PF$
Please refer to the figure for better understanding. Our expression will now have the following form
$\dfrac{PC-PB}{PB'-PC}=\dfrac{PF}{PB'-PF}$
As discussed above, we know the following information
$\begin{align}
  & u=PB \\
 & v=-B'P \\
 & f=-PF \\
 & R=-PC \\
\end{align}$
Substituting these values, we get
$\begin{align}
  & \dfrac{(-R-(-u))}{(-v-(-R)}=\dfrac{-f}{(-v-(-f))} \\
 & \Rightarrow \dfrac{u-R}{R-v}=\dfrac{-f}{f-v} \\
\end{align}$
Now we all know that the radius of curvature is twice the focal length, that is $R=2f$
We can now write our expression as
$\begin{align}
  & \dfrac{u-2f}{2f-v}=\dfrac{-f}{f-v} \\
 & \Rightarrow (u-2f)(f-v)=(-f)(2f-v) \\
 & \Rightarrow uf-uv-2{{f}^{2}}+2fv=-2{{f}^{2}}+fv \\
 & \Rightarrow f(u+v)=uv \\
\end{align}$
Taking reciprocal of both sides of the equation, we get
$\begin{align}
  & \dfrac{1}{f}=\dfrac{u+v}{uv} \\
 & \Rightarrow \dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v} \\
\end{align}$
Thus we have successfully derived our required equation.

Note
There were some assumptions we made that were crucial to our derivation, those assumptions are listed below. The distances we measured are from the pole of the mirror P. The distances below the axis are taken as negative and the distances above the axis are taken as positive. The light ray is travelling from left to right and we have used our sign convention accordingly in the solution.