Two Plano convex lenses of focal lengths ${f_1}$ and ${f_2}$ set for minimising spherical aberration are kept apart by $2cm$. If the equivalent focal length of combination is $10cm$, the values of ${f_1}$ and ${f_2}$are :
(A) $18cm,16cm$
(B) $18cm,20cm$
(C) $20cm,22cm$
(D) $14cm,20cm$
Answer
269.4k+ views
Hint: In order to solve this question, we will first find the value of ${f_2}$ in terms of ${f_1}$ and then by using the general lens combination formula, we will solve for the value of both the focal length of two Plano-convex lenses.
Formula used:
If ${f_1}$ and ${f_2}$ are the focal length of two lenses and they are kept at a distance of d then their combined focal length is calculated using formula $\dfrac{1}{{{f_{net}}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} - \dfrac{d}{{{f_1}{f_2}}}$
Complete answer:
We have given that two Plano convex lenses has a focal length of ${f_1}$ and ${f_2}$,and they are kept at a distance of $d = 2cm$ so, for minimum spherical aberration we have
$
{f_2} - {f_1} = 2 \\
\Rightarrow {f_2} = 2 + {f_1} \\
$
Now, using the formula $\dfrac{1}{{{f_{net}}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} - \dfrac{d}{{{f_1}{f_2}}}$ where ${f_{net}} = 10cm$ is given to us and solving for ${f_1}$ we get,
$
\dfrac{1}{{10}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{(2 + {f_1})}} - \dfrac{2}{{{f_1}(2 + {f_1})}} \\
\dfrac{1}{{10}} = \dfrac{{2 + {f_1} + {f_1} - 2}}{{{f_1}(2 + {f_1})}} \\
\dfrac{1}{{10}} = \dfrac{2}{{(2 + {f_1})}} \\
$
Rearranging the above equation we get,
$
2 + {f_1} = 20 \\
\Rightarrow {f_1} = 18cm \\
$
and from the relation ${f_2} = 2 + {f_1}$ we get,
$
{f_2} = 18 + 2 \\
{f_2} = 20cm \\
$
So, The focal length of two Plano convex lenses are
$
{f_1} = 18cm \\
{f_2} = 20cm \\
$
Hence, the correct option is (B) $18cm,20cm$
Note: It should be remembered that in the case of minimum aberration the difference in the focal length of Plano-convex lenses is equal to the actual distance between them and the SI unit of focal length is meter whereas the inverse of focal length is known as Power whose SI unit is known as Dioptre.
Formula used:
If ${f_1}$ and ${f_2}$ are the focal length of two lenses and they are kept at a distance of d then their combined focal length is calculated using formula $\dfrac{1}{{{f_{net}}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} - \dfrac{d}{{{f_1}{f_2}}}$
Complete answer:
We have given that two Plano convex lenses has a focal length of ${f_1}$ and ${f_2}$,and they are kept at a distance of $d = 2cm$ so, for minimum spherical aberration we have
$
{f_2} - {f_1} = 2 \\
\Rightarrow {f_2} = 2 + {f_1} \\
$
Now, using the formula $\dfrac{1}{{{f_{net}}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}} - \dfrac{d}{{{f_1}{f_2}}}$ where ${f_{net}} = 10cm$ is given to us and solving for ${f_1}$ we get,
$
\dfrac{1}{{10}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{(2 + {f_1})}} - \dfrac{2}{{{f_1}(2 + {f_1})}} \\
\dfrac{1}{{10}} = \dfrac{{2 + {f_1} + {f_1} - 2}}{{{f_1}(2 + {f_1})}} \\
\dfrac{1}{{10}} = \dfrac{2}{{(2 + {f_1})}} \\
$
Rearranging the above equation we get,
$
2 + {f_1} = 20 \\
\Rightarrow {f_1} = 18cm \\
$
and from the relation ${f_2} = 2 + {f_1}$ we get,
$
{f_2} = 18 + 2 \\
{f_2} = 20cm \\
$
So, The focal length of two Plano convex lenses are
$
{f_1} = 18cm \\
{f_2} = 20cm \\
$
Hence, the correct option is (B) $18cm,20cm$
Note: It should be remembered that in the case of minimum aberration the difference in the focal length of Plano-convex lenses is equal to the actual distance between them and the SI unit of focal length is meter whereas the inverse of focal length is known as Power whose SI unit is known as Dioptre.
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