Answer:
This the power of combination is given by
[tex]P_{combination}=\frac{1}{F}=\frac{d+f_{2}-f_{1}}{f_{2}(d-f_{1})}[/tex]
Explanation:
Let 'F' be the focal length of the combination of the two lenses and the focal length's of individual lenses be [tex]f_{1},f_{2}[/tex]
We know that focal length is the position of image when object is placed at infinity
Let us place the the object at infinity with respect to the first lens thus the position of image formed by the first lens shall be obtained using lens formula as
[tex]\frac{1}{f_{1}}=\frac{1}{u}+\frac{1}{v}[/tex]
Applying values we get
[tex]\frac{1}{f_{1}}=\frac{1}{\infty }+\frac{1}{v}\\\\\Rightarrow \frac{1}{f}=0+\frac{1}{v}\\\\\therefore v=+f_{1}[/tex]
Now this position of image formed by the first lens act's as object for the second lens, thus we have
[tex]\frac{1}{f_{2}}=\frac{1}{-(d-f_{1})}+\frac{1}{v}\\\\\Rightarrow \frac{1}{f_{2}}+\frac{1}{d-f_{1}}=\frac{1}{v}\\\\\therefore \frac{1}{v}=\frac{(d-f_{1})+f_{2}}{f_{2}(d-f_{1})}\\\\=\frac{1}{v}=\frac{d+f_{2}-f_{1}}{f_{2}(d-f_{1})}\\\\\therefore v_{f}=\frac{f_{2}(d-f_{1})}{d+f_{2}-f_{1}}[/tex]
Since image of an object placed at infinity will be formed at [tex]v_{f}=\frac{f_{2}(d-f_{1})}{d+f_{2}-f_{1}}[/tex] thus the focal length of the combination of the 2 thin lenses will be [tex]F=\frac{f_{2}(d-f_{1})}{d+f_{2}-f_{1}}[/tex]
This the power of combination is given by
[tex]P_{combination}=\frac{1}{F}=\frac{d+f_{2}-f_{1}}{f_{2}(d-f_{1})}[/tex]
(For the sake of question we assume lenses to be convex although the same procedure is valid for all other lenses)
