Drive an expression for equivalent focal length of two thin lenses kept in contact.

Question

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Answer 1

Let the focal lengths of the two lenses be $f_{1}$ and $f_{2}$ .
$u_{1} = O C_{1}$ and
$v_{1} = I_{1} C_{1}$
$u_{2} = - C_{1} I_{1}$ which is approximately equal to $C_{2} I_{1}$ and
$v_{2} = C_{2} I$ which is approximately equal to $C_{1} I$
$∴ \frac{1}{u _{1}} + \frac{1}{v _{1}} = \frac{1}{f _{1}}$
and $\frac{1}{u _{2}} + \frac{1}{v _{2}} = \frac{1}{f _{2}}$
$\frac{1}{O C _{1}} + \frac{1}{I _{1} C _{1}} = \frac{1}{f 1}$ and
$\frac{1}{- C _{1} I _{1}} + \frac{1}{C _{1} I} = \frac{1}{f _{2}}$
So $\frac{1}{O C _{1}} + \frac{1}{I _{1} C _{1}} = \frac{1}{f _{1}} + \frac{1}{f _{2}} = \frac{1}{F}$ .
Combined focal length of two thin lenses in contact is given by:
$F = \frac{f _{1} f _{2}}{f _{1} + f _{2}}$

Answer 2

Let the focal lengths of the two lenses be

f_{1}

and

f_{2}

.

u_{1} = O C_{1}

and

v_{1} = I_{1} C_{1}

u_{2} = - C_{1} I_{1}

which is approximately equal to

C_{2} I_{1}

and

v_{2} = C_{2} I

which is approximately equal to

C_{1} I

∴ \frac{1}{u _{1}} + \frac{1}{v _{1}} = \frac{1}{f _{1}}

and

\frac{1}{u _{2}} + \frac{1}{v _{2}} = \frac{1}{f _{2}}

\frac{1}{O C _{1}} + \frac{1}{I _{1} C _{1}} = \frac{1}{f 1}

and

\frac{1}{- C _{1} I _{1}} + \frac{1}{C _{1} I} = \frac{1}{f _{2}}

So

\frac{1}{O C _{1}} + \frac{1}{I _{1} C _{1}} = \frac{1}{f _{1}} + \frac{1}{f _{2}} = \frac{1}{F}

.
Combined focal length of two thin lenses in contact is given by:

F = \frac{f _{1} f _{2}}{f _{1} + f _{2}}