Consider a thin lens placed between a source (S) and an observer (O) (Fig). Let the thickness of the lens vary as $\text{w}\text{(b)}=\text{w}_0-\frac{\text{b}^2}{\alpha}$, where b is the verticle distance from the pole. $w_0$ is a constant. Using Fermat’s principle i.e. the time of transit for a ray between the source and observer is an extremum, find the condition that all paraxial rays starting from the source will converge at a point O on the axis. Find the focal length.
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The time taken by ray to travel from S to $P_1$ is
$\text{t}_1=\frac{\text{SP}_1}{\text{c}}=\frac{\sqrt{\text{u}^2+\text{b}^2}}{\text{c}}$
or $\text{t}_1=\frac{\text{u}}{\text{c}}\Big(1+\frac{1}{2}\frac{\text{b}^2}{\text{u}^2}\Big)$ assuming b < < 1.
The time required to travel from $P_1$ to O is
$\text{t}_2=\frac{\text{P}_1\text{O}}{\text{c}}=\frac{\sqrt{\text{v}^2+\text{b}^2}}{\text{c}}=\frac{\text{v}}{\text{c}}\Big(1+\frac{1}{2}\frac{\text{b}^2}{\text{v}^2}\Big)$
The time required to travel through the lens is
$\text{t}=\frac{(\text{n}-1)\text{w}(\text{b})}{\text{c}}$
where n is the refractive index.
Thus, the total time is
$\text{t}=\frac{1}{\text{c}}\bigg[\text{u}+\text{v}+\frac{1}{2}\text{b}^2\Big(\frac{1}{\text{u}}+\frac{1}{\text{v}}\Big)+(\text{n}-1)\text{w}(\text{b})\bigg]$
Put $\frac{1}{\text{D}}=\frac{1}{\text{u}}+\frac{1}{\text{v}}$
Then, $\text{t}=\frac{1}{\text{c}}\bigg[\text{u}+\text{v}+\frac{1}{2}\frac{\text{b}^2}{\text{D}}+(\text{n}-1)\Big(\text{w}_0+\frac{\text{b}^2}{\alpha}\Big)\bigg]$
Fermat's principle gives the time taken should be minimum.
For that first dericative should be zero.
$\frac{\text{dt}}{\text{db}}=0=\frac{\text{b}}{\text{CD}}-\frac{2(\text{n}-1)\text{b}}{\text{c}\alpha}$
Thus, a convergent lens is formed if $\alpha=2(\text{n}-1)\text{D}$. This is independent of and hence all paraxial rays from S will converge at O i.e., for rays b < < n and b < < v.
Since, $\frac{1}{\text{D}}=\frac{1}{\text{u}}+\frac{1}{\text{v}}$, the focal length is D.
$\text{t}_1=\frac{\text{SP}_1}{\text{c}}=\frac{\sqrt{\text{u}^2+\text{b}^2}}{\text{c}}$
or $\text{t}_1=\frac{\text{u}}{\text{c}}\Big(1+\frac{1}{2}\frac{\text{b}^2}{\text{u}^2}\Big)$ assuming b < < 1.
The time required to travel from $P_1$ to O is
$\text{t}_2=\frac{\text{P}_1\text{O}}{\text{c}}=\frac{\sqrt{\text{v}^2+\text{b}^2}}{\text{c}}=\frac{\text{v}}{\text{c}}\Big(1+\frac{1}{2}\frac{\text{b}^2}{\text{v}^2}\Big)$
The time required to travel through the lens is
$\text{t}=\frac{(\text{n}-1)\text{w}(\text{b})}{\text{c}}$
where n is the refractive index.
Thus, the total time is
$\text{t}=\frac{1}{\text{c}}\bigg[\text{u}+\text{v}+\frac{1}{2}\text{b}^2\Big(\frac{1}{\text{u}}+\frac{1}{\text{v}}\Big)+(\text{n}-1)\text{w}(\text{b})\bigg]$
Put $\frac{1}{\text{D}}=\frac{1}{\text{u}}+\frac{1}{\text{v}}$
Then, $\text{t}=\frac{1}{\text{c}}\bigg[\text{u}+\text{v}+\frac{1}{2}\frac{\text{b}^2}{\text{D}}+(\text{n}-1)\Big(\text{w}_0+\frac{\text{b}^2}{\alpha}\Big)\bigg]$
Fermat's principle gives the time taken should be minimum.
For that first dericative should be zero.
$\frac{\text{dt}}{\text{db}}=0=\frac{\text{b}}{\text{CD}}-\frac{2(\text{n}-1)\text{b}}{\text{c}\alpha}$
Thus, a convergent lens is formed if $\alpha=2(\text{n}-1)\text{D}$. This is independent of and hence all paraxial rays from S will converge at O i.e., for rays b < < n and b < < v.
Since, $\frac{1}{\text{D}}=\frac{1}{\text{u}}+\frac{1}{\text{v}}$, the focal length is D.
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