Question

Two similar thin $equi-convex$ lenses, of focal length $f$ each, are kept coaxially in contact with each other such that the focal length of the combination is $F_1$. When the space between the two lenses is filled with glycerin (which has the same refractive index $(p = 1.5)$ as that of glass) then the equivalent focal length is $F_2$. The ratio $F_1$ : $F_2$ will be :

• A. 2 : 3
• B. 3:4
• C. 2 : 1
• D. 1:2

Answer 1

The Correct Option is D

To solve this problem, we need to analyze the behavior of thin lenses in contact and then when a medium fills the space between them.

Step 1: Focal Length of Two Lenses in Contact

For two thin lenses in contact, the equivalent focal length $F_1$ is given by:

\(\frac{1}{F_1} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f}\).

Thus, $F_1 = \frac{f}{2}$.

Step 2: Equivalent Focal Length When Filled with Glycerin

When the space between the lenses is filled with glycerin having the same refractive index as the lenses, the system now behaves as a single medium with zero refractive power at the interface. Therefore, it acts like a single lens with one focal length.

The medium between them cancels the refractive power of surfaces between the lenses. Hence, the lenses together effectively act as one thick lens of combined focal length:

\(\frac{1}{F_2} = \frac{1}{f} + 0 + \frac{1}{f} = \frac{2}{f}\).

Thus, $F_2 = f$.

Step 3: Ratio of the Focal Lengths

Now, we determine the ratio $F_1 : F_2$:

\(\frac{F_1}{F_2} = \frac{\frac{f}{2}}{f} = \frac{1}{2}\).

The ratio $F_1 : F_2 = 1 : 2$.

Conclusion

Therefore, the correct answer is $1 : 2$.