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Show that, two thin lenses kept in contact, form an achromatic doublet if they satisfy the condition:
$\dfrac{\omega }{f} + \dfrac{{\omega '}}{{f'}} = 0$
where the terms have their usual meaning.

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Last updated date: 20th Jun 2024
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Answer
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Hint: We know that a single lens will have different focal lengths for different colors and the image formed by a single lens faces a problem named chromatic aberration. Now, by keeping two thin lenses in contact, we can remove chromatic aberration and this condition is known as achromatic combination or doublet.

Formula used:
Lens maker’s formula states that
$\dfrac{1}{{{f_v}}} = \left( {{\mu _v} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Dispersive power,
\[\omega = \dfrac{{\left( {{\mu _v} - {\mu _r}} \right)}}{{\left( {\mu - 1} \right)}}\]

Complete step by step solution:
Now, we know that by keeping two thin lenses in contact, we can remove chromatic aberration and this condition is known as achromatic combination or doublet.
Now, in achromatic combination focal lengths of the lens for colour violet and red are equal, ${F_v} = {F_r}$.
Now, we will use the lens maker’s formula for both the lenses.
So,
$\dfrac{1}{{{f_v}}} = \left( {{\mu _v} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ and
$\dfrac{1}{{{f_r}}} = \left( {{\mu _r} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Now, subtracting equation the above equations we get,

$\dfrac{1}{{{f_v}}} - \dfrac{1}{{{f_r}}} = \left( {{\mu _v} - {\mu _r}} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)..............\left( 1 \right)$
Now, if the mean focal length is $f$ , then

$\dfrac{1}{f} = \left( {\mu - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ so, it can also be written as,
$\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}} = \dfrac{1}{{{f_v}}}\left( {{\mu _v} - 1} \right)$
Now, substituting this value in equation (1),
We get,

$\dfrac{1}{{{f_v}}} - \dfrac{1}{{{f_r}}} = \dfrac{{\left( {{\mu _v} - {\mu _r}} \right)}}{{\left( {\mu - 1} \right)f}}................\left( 2 \right)$
Now, as we know,
$\dfrac{{\left( {{\mu _v} - {\mu _r}} \right)}}{{\left( {\mu - 1} \right)}} = \omega $
Now, substituting this in the equation (2),
$\dfrac{1}{{{f_v}}} - \dfrac{1}{{{f_r}}} = \dfrac{\omega }{f}............\left( 3 \right)$
Now, similarly, we can write the above equation for the second lens of dispersive power $\omega '$ and focal length $f'$ ,
So,
$\dfrac{1}{{{{f'}_v}}} - \dfrac{1}{{{{f'}_r}}} = \dfrac{{\omega '}}{{f'}}............\left( 4 \right)$
Now, focal lengths for violet and red color will be \[{F_v}\] and \[{F_r}\],
Now, we can write that,

\[\dfrac{1}{{{F_v}}} = \dfrac{1}{{{f_v}}} + \dfrac{1}{{{{f'}_v}}}\]
And
\[\dfrac{1}{{{F_r}}} = \dfrac{1}{{{f_r}}} + \dfrac{1}{{{{f'}_r}}}\]
Now, we know that for an achromatic combination, ${F_v} = {F_r}$ .
So,
 \[
  \dfrac{1}{{{f_v}}} + \dfrac{1}{{{{f'}_v}}} = \dfrac{1}{{{f_r}}} + \dfrac{1}{{{{f'}_r}}} \\
  \left( {\dfrac{1}{{{f_v}}} - \dfrac{1}{{{f_r}}}} \right) + \left( {\dfrac{1}{{{{f'}_v}}} - \dfrac{1}{{{{f'}_r}}}} \right) = 0.................\left( 5 \right) \\
 \]

Now, putting equation (3) and (4) in equation (5)
We get,
$\dfrac{\omega }{f} + \dfrac{{\omega '}}{{f'}} = 0$ ,
which is the required condition.

Note: As we know, we have to use the lens maker’s formula to solve this question. Along with this, we have to be clear about the conditions of chromatic aberration and achromatic combination. Also, the above question is a bit typical. So, try to be cautious in doing the calculations.