Solveeit Logo

Question

Question: A convergent doublet of separated lenses, corrected for spherical aberration, has a resultant focal ...

A convergent doublet of separated lenses, corrected for spherical aberration, has a resultant focal length of 10 cm. The separation between the two lenses is 2 cm. The focal lengths of the component lenses are
A) 18cm,20cmA)\text{ }18cm,20cm
B) 10cm,12cmB)\text{ }10cm,12cm
C) 12cm,14cmC)\text{ }12cm,14cm
D) 16cm,18cmD)\text{ }16cm,18cm

Explanation

Solution

This problem can be solved by finding out the focal lengths of the combination of the two lenses and equating it to the resultant of the focal length of a doublet corrected for spherical aberration in terms of the focal lengths of the two lenses and the distance between them. We will get a required condition for the focal lengths of the lenses which should be satisfied by one of the options.

Formula used:
1f=f1+f2df1f2\dfrac{1}{f}=\dfrac{{{f}_{1}}+{{f}_{2}}-d}{{{f}_{1}}{{f}_{2}}}

Complete step-by-step answer:
We will equate the equivalent focal length of the combination of lenses with the resultant focal length of the doublet of lenses corrected for spherical aberration.
Also, the resultant focal length ff of a doublet of lenses corrected for spherical aberration is given by
1f=f1+f2df1f2\dfrac{1}{f}=\dfrac{{{f}_{1}}+{{f}_{2}}-d}{{{f}_{1}}{{f}_{2}}} --(1)
Where f1,f2{{f}_{1}},{{f}_{2}} are the focal lengths of the two lenses and dd is the distance of separation between them.
Now, let us analyze the question.
Let the focal length of the two lenses be f1,f2{{f}_{1}},{{f}_{2}} respectively.
The distance between them is d=10cmd=10cm.
It is also given that the equivalent focal length of the combination is feq=10cm{{f}_{eq}}=10cm.
Therefore, using (1), we get
1feq=f1+f2df1f2\dfrac{1}{{{f}_{eq}}}=\dfrac{{{f}_{1}}+{{f}_{2}}-d}{{{f}_{1}}{{f}_{2}}}
110=f1+f22f1f2\Rightarrow \dfrac{1}{10}=\dfrac{{{f}_{1}}+{{f}_{2}}-2}{{{f}_{1}}{{f}_{2}}} --(2)
Now, we will check the options one by one to see which one satisfies the above condition.
A) 18cm,20cmA)\text{ }18cm,20cm
Putting in (2), we get
110=18+20218×20=3618×20=110\Rightarrow \dfrac{1}{10}=\dfrac{18+20-2}{18\times 20}=\dfrac{36}{18\times 20}=\dfrac{1}{10}
which is true.
Therefore, this set of focal lengths is the required focal lengths of the two spherical lenses.
Therefore, the correct option is A) 18cm,20cmA)\text{ }18cm,20cm.

So, the correct answer is “Option A”.

Note: The familiar formula for the focal length of a combination of lenses, that is,
1feq=i=1n1fi\dfrac{1}{{{f}_{eq}}}=\sum\limits_{i=1}^{n}{\dfrac{1}{{{f}_{i}}}}
Is only valid when the separation between the lenses is zero, that is the lenses are placed touching each other. Since, in this question, that is not the case, this formula is not applicable here. Many students might make the mistake of using this formula. In fact, the more general formula is (1), and actually it becomes the familiar formula for the equivalent focal length of a combination of lenses touching each other, when the separation dd is deemed to be zero in the equation.