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 (\(\mu=-1.5\)) as that of glass) then the equivalent focal length is\( F_2.\) The ratio \(F1: F2\) will be: 

• A. \(2:1\)
• B. \(1:2\)
• C. \(2:3\)
• D. \(3:4\)

Answer 1

The Correct Option is B

To solve this problem, we need to analyze the optical system consisting of two equi-convex lenses. Both lenses have the same focal length \( f \), and they are placed coaxially in contact. The given refractive index of the lens material (glass) and the liquid (glycerin) filling the space is \(\mu = 1.5\).

Step-by-Step Solution:

1. According to the lens maker's formula, the focal length \( F_1 \) of two similar thin lenses in contact is given by: \( \frac{1}{F_1} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f} \). Thus, the resulting focal length when the lenses are in contact is \( F_1 = \frac{f}{2} \).
2. When a converging lens is filled with a medium of the same refractive index, it effectively nullifies the optical effect of the surfaces that are in contact with the medium. Hence, the glycerin will essentially negate the effect of the lens surfaces that touch the liquid.
3. Consequently, when the two lenses are separated by glycerin, the liquid does not alter the optical properties further, as glycerin has the same refractive index as that of glass. Thus, the effective focal length of the entire system remains that of a single lens with focal length \( f \): \( F_2 = f \).
4. Now, we calculate the ratio \( F_1 : F_2\): \( \frac{F_1}{F_2} = \frac{\frac{f}{2}}{f} = \frac{1}{2} \). Therefore, the ratio \( F_1 : F_2 = 1:2 \).

Conclusion: Since the equivalent focal length \( F_2 \) when glycerine fills the space becomes \( f \), while it was \( \frac{f}{2} \) when lenses were in contact, the ratio \( F_1 : F_2 \) is \( 1:2 \).