Question 231: Ray Optics

Question

A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of 10 cm. The separation between the two lenses is $$2$$ cm. The focal lengths of the component lenses are :

Accepted Answer

For a convergent doublet of separated lens, we have
$${1 \over f} = {1 \over {{f_1}}} + {1 \over {{f_2}}} - {d \over {{f_1}{f_2}}}$$ ...... (1)
where d is separation between two lens, f
1
and f
2
are focal lengths of component lenses, f is resultant focal length. Therefore, Eq. (1) becomes
$${1 \over {10}} = {1 \over {{f_1}}} + {1 \over {{f_2}}} - {2 \over {{f_1}{f_2}}} \Rightarrow {1 \over {10}} = \left( {{{{f_2} + {f_1} - 2} \over {{f_1}{f_2}}}} \right)$$
$$ \Rightarrow {f_1}{f_2} = 10{f_2} + 10{f_1} - 20$$
$$ \Rightarrow 10{f_1} + 10{f_2} - {f_1}{f_2} =  + 20$$
For f
1
= 18 cm and f
2
= 20 cm, the above equation satisfies.

Answer

10 cm, 12 cm

Answer

12 cm, 14 cm

Answer

16 cm, 18 cm

Answer

18 cm, 20cm

Ray Optics

A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of 10 cm. The separation between the two lenses is $2$ cm. The focal lengths of the component lenses are :

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