Question

Two identical double convex lenses each of focal length f are made of glass $\mu =\frac{3}{2}.$ If R is the radius of curvature of each of the four faces, then

• A. The focal length of the combination is half the radius of curvature of any of the four surfaces.
• B. The focal length of the combination increases if the lenses are separated by a distance f and the new focal length is twice the focal length when the lenses are in contact.
• C. The focal length of the combination decreases when they are separated by a distance f and the new focal length is half the focal length when the lenses are in contact.
• D. None of these.

Answer 1

Answer: (A), (B)

The focal length of the combination is half the radius of curvature of any of the four surfaces.
The focal length of the combination increases if the lenses are separated by a distance f and the new focal length is twice the focal length when the lenses are in contact.

$\frac{1}{f}=(\mu -1)(\frac{1}{R_1}+\frac{1}{R_2})$

$=(\frac{3}{2}-1)(\frac{1}{R}+\frac{1}{R})=\frac{1}{R}$

$\therefore f=R$

Combined focal length when in contact

$\frac{1}{F}=\frac{1}{f}+\frac{1}{f}=\frac{2}{f}=\frac{2}{R}$

$\therefore F=\frac{R}{2}$

Combined focal length when they are separated by $f$,

$\frac{1}{F^{\prime }}=\frac{1}{f}+\frac{1}{f}-\frac{f}{f^2}$

$\frac{1}{F^{\prime }}=\frac{1}{f}$

$F^{\prime }=f=R$

$\therefore F^{\prime }=2F$