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  • (i) Consider a thin lens placed between a source (S) and an observer (O) (Fig. 9.8). Let the thickness of the lens vary as w(b)=w_{o}-frac{b_{2}}{alpha } where b is the vertical distance from the pole. w_{o} is a constant.

(i) Consider a thin lens placed between a source (S) and an observer (O) (Fig. 9.8). Let the thickness of the lens vary as w(b)=w_{o}-\frac{b_{2}}{\alpha } where b is the vertical distance from the pole. w_{o} 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.

(ii) A gravitational lens may be assumed to have a varying width of the form

w(b)=k_{1}ln\left ( \frac{k_{2}}{b} \right ) b_{min}<b<b_{max}

=k_{1}ln\left ( \frac{k_{2}}{b_{min}} \right )b<b_{min}

Show that an observer will see an image of a point object as a ring about the center of the lens with an angular radius

\beta =\sqrt{\frac{\left ( n-1 \right )k_{1}\frac{u}{v}}{u+v}}

 

Answers (1)

(i) The time taken by ray to travel from S to P_{1} is

             t_{1}=\frac{SP_{1}}{c}=\frac{\sqrt{u^{2}+b^{2}}}{c}

       Or t_{1}=\frac{u}{c}\left ( 1+\frac{1}{2}\frac{b^{2}}{u^{2}} \right ) assuming b<<1

The time required to travel from P_{1} to O is

t_{2}=\frac{P_{1}O}{c}=\frac{\sqrt{v^{2}+b^{2}}}{c}=\frac{v}{c}\left ( 1+\frac{1}{2}\frac{b^{2}}{v^{2}} \right )

The time required to travel through the lens is

t_{l}=\frac{(n-1)w(b)}{c}

where n is the refractive index.

Thus, the total time is

t=\frac{1}{c}\left [ u+v+\frac{1}{2}b^{2}\left ( \frac{1}{u}+\frac{1}{v} \right )+(n-1)w(b) \right ]

Put \frac{1}{D}=\frac{1}{u}+\frac{1}{v}

Then,

t=\frac{1}{c}\left ( u+v+\frac{1}{2}\frac{b^{2}}{D}+\left ( n-1 \right )\left ( w_{0}+\frac{b^{2}}{\alpha } \right ) \right )

Fermat's principle gives the time taken should be minimum. For that first derivative should be zero.

\frac{dt}{db}=0=\frac{b}{CD}-\frac{2(n-1)b}{c\alpha }

      \alpha =2(n-1)D

Thus, a convergent lens is formed if \alpha =2(n-1)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}{D}=\frac{1}{u}+\frac{1}{v}, the focal length is D.

(ii) In this case, differentiating expression of time taken t e.r.t. b.

        t=\frac{1}{c}\left ( u+v+\frac{1}{2}\frac{b^{2}}{D}+(n-1)k_{1}ln\left ( \frac{k_{2}}{b} \right ) \right )

       \frac{dt}{db}=0=\frac{b}{D}-(n-1)\frac{k_{1}}{b}

\Rightarrow b^{2}=(n-1)k_{1}D

\therefore b=\sqrt{(n-1)k_{1}D}

Thus, all rays passing at a height b shall contribute to the image. The ray paths make an angle.

\beta =\frac{b}{v}=\frac{\sqrt{(n-1)k_{1}D}}{v}=\sqrt{\frac{(n-1)k_{1}uv}{v^{2}(u+v)}}=\sqrt{\frac{(n-1)k_{1}u}{(u+v)v}} This is the required expression.

Posted by

infoexpert23

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