
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.
Answer
173.4k+ views
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.
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.
Recently Updated Pages
JEE Main Mock Test 2025-26: Chapter-Wise Practice Papers

JEE Main Electromagnetic Waves Mock Test 2025-26 | Free Practice Online

JEE Main 2025-26 Electronic Devices Mock Test: Free Practice Online

JEE Main Mock Test 2025-26: Current Electricity Practice Online

JEE Main 2025-26 Electrostatics Mock Test – Free Practice Online

JEE Main 2025-26 Units and Measurements Mock Test Online

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

Electric field due to uniformly charged sphere class 12 physics JEE_Main

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Instantaneous Velocity - Formula based Examples for JEE

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Essential Derivations for CBSE Class 12 Physics: Stepwise & PDF Solutions

Electron Gain Enthalpy and Electron Affinity for JEE

Wheatstone Bridge for JEE Main Physics 2025
