Question

Now if a concave lens of very large focal length $f_2$ is placed in contact with the convex lens then the shifting of the image is

• A. $\frac{f_1^2}{f_2}$
• B. $\frac{4f_1^2}{f_2}$
• C. $\frac{8f_1^2}{f_2}$
• D. $\frac{f_1^2}{4f_2}$

Answer 1

The correct option is B $\frac{4f_1^2}{f_2}$

Due to combination of the two lenses, focal length of the combination,F is given by

$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$

$u=-2f_1$

Thererfore, $\frac{1}{v}-\frac{1}{-2f_1}=\frac{1}{f_1}+\frac{1}{f_2}$

Hence $v=\frac{2f_1{f}_2}{2f_1+f_2}$

Previously the image was at the $C_2$

The shift in the image position=$\frac{2f_1{f}_2}{2f_1+f_2}-2f_1=\frac{-4f_1^2}{2f_1+f_2}$

Since $f_2$ is large, $2f_1+f_2\approx f_2$

Thus the shift is $\frac{-4f_1^2}{f_2}$($f_2$ is negative).