The focal length of a combination of two lenses is doubled if the separation between them is doubled. If the separation is increased to 4 times, the magnitude of focal length is

Options

A. doubled
B. quadrupled
C. halved
D. same

Answer

A

Explanation

`1/F=1/(f_1)+1/(f_2)-d/(f_1f_2)`
`=(f_1+f_2)/(f_1f_2)-d/(f_1f_2)`
`1/F=((f_1+f_2)-d)/(f_1f_2)`
`:. F= (f_1f_2)/((f_1+f_2)-d)…..(i)`
If d is doubled, focal length is doubled or
denominator becomes half.
`:. (f_1+f_2)-2d=1/2[(f_1+f_2)-d]`
or `(f_1+f_2)=3d`
Substituting in Eq.(i), we have
Originally,
`F=(f_1f_2)/((f_1+f_2)-d)=(f_1f_2)/(3d-d)=(f_1f_2)/(2d)`
When d is made 4 times.
`F'=(f_1f_2)/((f_1+f_2)-4d)=(f_1f_2)/(3d-4d)=(f_1f_2)/d`
`=-2F`