Question

Two identical glass $({\mu }_g=3/2)$ equiconvex lenses of focal length $f$ each are kept in contact. The space between the two lenses is filled with water $({\mu }_w=4/3).$ The focal length of the combination is

• A. $3f/4$
• B. $f/3$
• C. $f$
• D. $4f/3$

Answer 1

The correct option is A $3f/4$

Given, two identical glass equiconvex lens of focal length $f$ is are kept in contact, and refractive index of glass is ${\mu }_g=3/2$.

The space between the two lenses is filled with water of refractive index ${\mu }_w=4/3$.

Here, the water in between the lenses can be considered as a concave lens.

Then it can be said that the radius of curvature of all the lenses is same.

Now, if the focal length of the convex lenses is $f$, we have,

$\frac{1}{f}=\frac{({\mu }_g-1)2}{R}=\frac{(\frac{3}{2}-1)2}{R}=\frac{1}{R}$

Where, $R$ be the radius of curvature.

Thus, $f=R$.

Now, let, the focal length of concave lens is $f\prime$. Then, we have,

$\frac{1}{f\prime }=\frac{({\mu }_w-1)(-2)}{R}=\frac{-2}{3f}$

Now, if $F$ be the focal length of the combination, then,

$\frac{1}{F}=\frac{1}{f}+\frac{-2}{3f}+\frac{1}{f}=\frac{4}{3f}$

$\therefore F=\frac{3f}{4}$

Thus, the correct option is - (A) $\frac{3f}{4}$.