A convergent doublet of separated lens, corrected for spherical aberration, are separated by 2 cm and has an equivalent focal length of 10 cm. The focal length of its component lenses are:

Question

A convergent doublet of separated lens, corrected for spherical aberration, are separated by 2 cm and has an equivalent focal length of 10 cm. The focal length of its component lenses are: (A)  f1=18 cm, f2=10 cm  (B) f1=20 cm , f2=28 cm (C) f1=20 cm , f2=18 cm (D) f1=24 cm , f2=18 cm

Tardigrade Admin · Accepted Answer

Since, the doublet is corrected for spherical.
aberration, it satisfies the following condition.

f_{1} - f_{2} = d = 2 c m

\Rightarrow f_{1} = f_{2} + 2 m

Let the equivalent focal length

= F : F

= \frac{f _{1} f _{2}}{f _{1} + f _{2} = d} = 10 c m

Solving it,

f_{1} = 20 c m, f_{2} = 18 c m .

Q. A convergent doublet of separated lens, corrected for spherical aberration, are separated by $2 c m$ and has an equivalent focal length of $10 c m$ . The focal length of its component lenses are:

$f_{1} = 18 c m, f_{2} = 10 c m$

$f_{1} = 20 c m, f_{2} = 28 c m$

$f_{1} = 20 c m, f_{2} = 18 c m$

$f_{1} = 24 c m, f_{2} = 18 c m$

Q. A convergent doublet of separated lens, corrected for spherical aberration, are separated by 2cm and has an equivalent focal length of 10cm. The focal length of its component lenses are:

f1​=18cm,f2​=10cm

f1​=20cm,f2​=28cm

f1​=20cm,f2​=18cm

f1​=24cm,f2​=18cm

Since, the doublet is corrected for spherical. aberration, it satisfies the following condition. f1​−f2​=d=2cm ⇒f1​=f2​+2m Let the equivalent focal length =F:F =f1​+f2​=df1​f2​​=10cm Solving it, f1​=20cm,f2​=18cm.

Q. A convergent doublet of separated lens, corrected for spherical aberration, are separated by $2 c m$ and has an equivalent focal length of $10 c m$ . The focal length of its component lenses are:

$f_{1} = 18 c m, f_{2} = 10 c m$

$f_{1} = 20 c m, f_{2} = 28 c m$

$f_{1} = 20 c m, f_{2} = 18 c m$

$f_{1} = 24 c m, f_{2} = 18 c m$