Derive n_1 -u +n_2 v =n_2-n_1 R for refraction at a spherical sur… - Vidyadip

Question

Derive $\frac{n_1}{-u}+\frac{n_2}{v}=\frac{n_2-n_1}{ R }$ for refraction at a spherical surface.

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Answer

→ As shown in figure, a point like object $O$ is placed on the principal axis of the spherical surface. A spherical surface has centre of curvature ' $C$ ' and radius of curvature $R$.
→ Rays emerge from a medium having refractive index $n_1$. Here, OM and ON are the incident rays.
→ They refract in a medium having refractive index $n_2$. Here NI and MI are the refractive rays. As a result, image I of the point object $O$ is obtained.
→ Assume that the aperture of the spherical surface is small compared to the object distance, image distance and radius of curvature, so that the angles can be taken small.
→ Since the aperture of the surface is assumed to be small here, NM will be taken to be nearly equal to the length of the perpendicular from the point N on the principal axis.
→ From figure,
$
\begin{array}{l}
\tan \angle NOM \approx \angle NOM=\frac{MN}{OM}...(1) \\
\tan \angle NCM \approx \angle NCM=\frac{MN}{MC} \ldots(2) \\
\tan \angle NIM \approx \angle NIM=\frac{MN}{MI} \ldots(3)
\end{array}
$
→ For $\triangle N O C, i$ is the exterior angle.
Therefore,
$
i=\angle NOM+\angle NCM
$
Substituting values from equation (1) and equation (2),
$
\therefore i=\frac{MN}{OM}+\frac{MN}{MC} \ldots(4)
$
→ From figure for $\triangle NIC , \angle NCM$ is the exterior angle.
$
\begin{array}{l}
\therefore \angle NCM=r+\angle NIM \\
r=\angle NCM-\angle NIM
\end{array}
$
$
\therefore r=\frac{MN}{MC}-\frac{MN}{MI} \ldots(5)
$
→ By applying Snell's law at point N ,
$
\begin{array}{l}
n_1 \sin i=n_2 \sin r \\
\text { But, } \sin i \approx i \\
\sin r \approx r \\
\therefore n_1 i=n_2 r
\end{array}
$
→ Substituting $i$ and $r$ from equation (4) and equation (5),
$
\begin{array}{l}
\therefore n_1\left(\frac{MN}{OM}+\frac{MN}{MC}\right)=n_2\left(\frac{MN}{MC}-\frac{MN}{MI}\right) \\
\therefore \frac{n_1}{OM}+\frac{n_1}{MC}=\frac{n_2}{MC}-\frac{n_2}{MI} \\
\therefore \frac{n_1}{OM}+\frac{n_2}{MI}=\frac{n_2}{MC}-\frac{n_1}{MC} \\
\therefore \frac{n_1}{OM}+\frac{n_2}{MI}=\frac{n_2-n_1}{MC}
\end{array}
$
→ But from figure, applying Cartesian sign convention,
$
\begin{array}{l}
OM=-u, MI=v \text { and } MC=R \\
\therefore-\frac{n_1}{u}+\frac{n_2}{v}=\frac{n_2-n_1}{R}
\end{array}
$
→ Above equation gives us a relation between object and image distance in terms of refractive index of the medium and the radius of curvature of the curved spherical surface.

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