Two similar thin equi-convex lenses, of focal 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 lens 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 `F_(1): F_(2)` will be

Options

A. `3:2`
B. `2:1`
C. `1:2`
D. `2:3`

Explanation

To solve the problem, we need to find the ratio of the focal lengths \( F_1 \) and \( F_2 \) for two configurations of the lenses. Let's break down the solution step by step.

### Step 1: Focal Length of Two Lenses in Contact

1. **Understanding the Configuration**: We have two similar thin equi-convex lenses in contact with each other, each having a focal length \( f \).

2. **Using the Lens Formula**: The formula for the focal length of two thin lenses in contact is given by:
\[ \frac{1}{F_1} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f} \]
Therefore, the focal length \( F_1 \) of the combination is:
\[ F_1 = \frac{f}{2} \]

### Step 2: Focal Length with Glycerin Between the Lenses

1. **Understanding the New Configuration**: When the space between the two lenses is filled with glycerin, which has the same refractive index (\( \mu = 1.5 \)) as the lenses, we can treat the system as three lenses: the two outer lenses (convex) and the middle lens (concave).

2. **Finding the Focal Length of the Middle Lens**: For the middle lens (glycerin), we use the lens formula:
\[ \frac{1}{f_{glycerin}} = \frac{\mu_{glycerin}}{\mu_{air} - 1} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Here, \( R_1 = -r \) (for the first surface) and \( R_2 = r \) (for the second surface). Thus:
\[ \frac{1}{f_{glycerin}} = \frac{1.5}{1 - 1} \left( \frac{1}{-r} - \frac{1}{r} \right) = \frac{1.5}{0} \left( -\frac{2}{r} \right) = -\frac{3}{r} \]
This indicates that the middle lens acts as a concave lens with focal length:
\[ f_{glycerin} = -\frac{r}{3} \]

3. **Finding the Total Focal Length \( F_2 \)**: Now, we can find the total focal length \( F_2 \) of the system:
\[ \frac{1}{F_2} = \frac{1}{f} + \frac{1}{f} + \frac{1}{f_{glycerin}} = \frac{1}{f} + \frac{1}{f} - \frac{3}{r} \]
Substituting \( f = \frac{r}{2} \):
\[ \frac{1}{F_2} = \frac{2}{\frac{r}{2}} - \frac{3}{r} = \frac{4}{r} - \frac{3}{r} = \frac{1}{r} \]
Therefore, the focal length \( F_2 \) is:
\[ F_2 = r \]

### Step 3: Finding the Ratio \( F_1 : F_2 \)

1. **Calculating the Ratio**:
\[ \frac{F_1}{F_2} = \frac{\frac{r}{2}}{r} = \frac{1}{2} \]

Thus, the ratio \( F_1 : F_2 \) is:
\[ \boxed{1 : 2} \]