Question

Derive the following expression for the refraction at concave spherical surface:
 ${\displaystyle \frac{\mu }{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{\mu -1}{R}}.$

Answer 1

Hint
A spherical mirror is a part of a sphere with a reflecting surface. If the inner surface is the reflective surface then the mirror is called a concave mirror and if the outer surface is the reflecting surface then the mirror is called a convex mirror. Here we have a concave reflecting surface for which we have to derive the given expression.

Complete step by step answer
Let us consider a concave mirror as shown in the diagram below,

The concave mirror is represented by $MPN$ . The refractive index of the medium of the spherical surface is given by $\mu$ . We consider $P$ as the pole of the mirror, $O$ as the centre of curvature, and $PC$ represents the principal axis of the refractive spherical surface. We consider a point object at $O$ . An incident ray travels through $C$ and it is normal to the spherical surface. It will not undergo any refraction hence it will travel in a straight line along $PX$ . Another ray that we consider is $OA$ . It will refract at the point $A$ bending towards the normal. At a point $i$ we will get a virtual image.
Let us consider the ray angles with the principal axis to be $\alpha ,\beta$ , and $\gamma$ respectively.
According to Snell’s law, we can write the refractive index as
 $\Rightarrow \mu ={\displaystyle \frac{\sin i}{\sin r}}$
Where $i$ is the angle of incidence and $r$ is the angle of refraction.
Here we have very small $i$ and $r$ , hence we can write,
 $\Rightarrow \sin i=i$ And $\sin r=r$ in equation
We get,
 $\Rightarrow \mu ={\displaystyle \frac{i}{r}}$
From this we get,
 $\Rightarrow i=\mu r$
By using the exterior angle theorem, from $\mathrm{\Delta }AOC$ we get,
 $\Rightarrow \gamma =i+\alpha$
From this we get
 $\Rightarrow i=\gamma -\alpha$
Now, by using exterior angle theorem in $\mathrm{\Delta }IAC$ , we get
 $\Rightarrow \gamma =\beta +r$
From this we get,
 $\Rightarrow r=\gamma -\beta$
Substituting these values of $i$ and $r$ in equation we get,
 $\Rightarrow \left(\gamma -\alpha \right)=\mu \left(\gamma -\beta \right)$
For a spherical surface, we can write the angle as
 $\Rightarrow angle={\displaystyle \frac{arc}{radius}}$
We can write
 $\Rightarrow \alpha ={\displaystyle \frac{PA}{OP}}$
And
 $\Rightarrow \beta ={\displaystyle \frac{PA}{IP}}$
Also
 $\Rightarrow \gamma ={\displaystyle \frac{PA}{CP}}$
Substituting these values of $\alpha ,\beta$ and $\gamma$ in , we get
 $\Rightarrow {\displaystyle \frac{PA}{PC}}-{\displaystyle \frac{PA}{PO}}=\mu \left({\displaystyle \frac{PA}{PC}}-{\displaystyle \frac{PA}{PI}}\right)$
Taking the common terms outside we get,
$\Rightarrow PA\left({\displaystyle \frac{1}{PC}}-{\displaystyle \frac{1}{PO}}\right)=\mu PA\left({\displaystyle \frac{1}{PC}}-{\displaystyle \frac{1}{PO}}\right)$
 $\Rightarrow PA$ gets cancelled as it is common on both sides. Now we have
 $\Rightarrow {\displaystyle \frac{1}{PC}}-{\displaystyle \frac{1}{PO}}=\mu \left({\displaystyle \frac{1}{PC}}-{\displaystyle \frac{1}{PI}}\right)$
Now we have to apply the sign convention.
 $\Rightarrow PC=-R$ (where $R$ is the radius of curvature)
 $\Rightarrow PI=-v$ (Where $v$ is the distance of the image from the pole of the mirror)
 $\Rightarrow PO=-u$ (Where $u$ is the distance of the object from the pole of the mirror)
Putting these values in equation
We get,
 $\Rightarrow \left({\displaystyle \frac{1}{-R}}\right)-\left({\displaystyle \frac{1}{-u}}\right)=\mu \left({\displaystyle \frac{1}{-R}}+{\displaystyle \frac{1}{v}}\right)$
Opening the brackets on LHS
$\Rightarrow {\displaystyle \frac{-1}{R}}+{\displaystyle \frac{1}{u}}=\mu \left({\displaystyle \frac{-1}{R}}+{\displaystyle \frac{1}{v}}\right)$
Opening the brackets on RHS
 $\Rightarrow {\displaystyle \frac{-1}{R}}+{\displaystyle \frac{1}{u}}={\displaystyle \frac{\mu }{R}}+{\displaystyle \frac{\mu }{v}}$
Rearranging, we get
 $\Rightarrow {\displaystyle \frac{-1}{R}}+{\displaystyle \frac{\mu }{R}}={\displaystyle \frac{\mu }{v}}-{\displaystyle \frac{1}{u}}$
We can write this expression as,
 $\Rightarrow {\displaystyle \frac{\mu -1}{R}}={\displaystyle \frac{\mu }{v}}-{\displaystyle \frac{1}{u}}$
Hence we got the required expression for the concave refractive surface.

Note
According to the Cartesian sign convention,
-All distances as measured from the pole of the mirror.
-The distances that are measured in the direction of the incident light are taken as positive.
-The distances that are measured opposite to the direction of incident light is considered as negative.
-The height measured upward the principal axis is measured as positive and the height measured downward is measured negative.