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(a) Using the ray diagram for a system of two lenses of focal lengths $f_{1}$ and $f_{2}$ in contact with each other, show that the two lens system can be regarded as equivalent to a single lens of focal length f, where $\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$ . Also write the relation for the equivalent power of the lens combination.

(b) Determine the position of the image formed by the lens combination given in the figure.

Answers (1)

(a) Let us consider the two thin lenses A and B forming the image $I$ and $I_{1}$ respectively, from the object (O) For I the distance is u and for $I_{1}$ the distance is $v_{1}$

For lens 'A', the lens formula is ;

$\frac{1}{f_{1}}=\frac{1}{v}-\frac{1}{u}$ _______(1)

Where v is the image distance

u is the object distance

For lens 'B' , the lens formula is :

$\frac{1}{f_{2}}=\frac{1}{v}-\frac{1}{v_{1}}$ ________(2)

On adding both equations, we get
$\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{v_{1}}-\frac{1}{u}+\frac{1}{v}-\frac{1}{v_{1}}$

$\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{v}-\frac{1}{u}$ $\left [ here, \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \right ]$

then,

$\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$

Therefore, the equivalent power form here would be

$P=P_{1}+P_{2}$

(b) The image formed by the lense of $f=+10cm$

$\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}$

$\frac{1}{v_{1}}-\frac{1}{30}=\frac{1}{10}$

$\therefore v_{1}=15cm$

This image formed by the lens act as an object from the concave lens.

Therefore,

$u_{2}=15-5=10cm$

$\frac{1}{f_{2}}+\frac{1}{v_{2}}=\frac{1}{u_{2}}$

$\frac{1}{-10}=\frac{1}{v}-\frac{1}{10}$

here, $v=\infty$

therefore, virtual image forms at the right of the concave lens at $v=\infty$ and act as an object for convex lens $(f=+30cm)$

$\frac{1}{v_{3}}-\frac{1}{\infty}=\frac{1}{f_{3}}$

$v_{3}=30cm$

Posted by

Safeer PP

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