Two glasses with refractive indices of `1.5 and 1.7` are used to make two indentical double convex lenses.
Find the ratio between their focal lengths.
How will each of these lenses act on a ray parallel to its optical axis if the lenses are submerged into a transparent liquid with a refractive index of `1.6?`

Explanation

To solve the problem step by step, we will first find the ratio of the focal lengths of the two lenses made from glasses with refractive indices of 1.5 and 1.7. Then, we will analyze how each lens behaves when submerged in a liquid with a refractive index of 1.6.

### Step 1: Finding the Focal Lengths

The formula for the focal length \( f \) of a double convex lens is given by:

\[ \frac{1}{f} = \left( n - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

For a double convex lens, if we assume \( R_1 = R \) and \( R_2 = -R \) (since the second radius is negative), the formula simplifies to:

\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R} + \frac{1}{R} \right) = (n - 1) \frac{2}{R} \]

### Step 2: Calculate the Focal Length for Glass with Refractive Index 1.5

For the first lens with a refractive index \( n_1 = 1.5 \):

\[ \frac{1}{f_1} = (1.5 - 1) \frac{2}{R} = 0.5 \frac{2}{R} = \frac{1}{R} \]

Thus,

\[ f_1 = R \]

### Step 3: Calculate the Focal Length for Glass with Refractive Index 1.7

For the second lens with a refractive index \( n_2 = 1.7 \):

\[ \frac{1}{f_2} = (1.7 - 1) \frac{2}{R} = 0.7 \frac{2}{R} = \frac{1.4}{R} \]

Thus,

\[ f_2 = \frac{R}{1.4} \]

### Step 4: Find the Ratio of the Focal Lengths

Now, we can find the ratio of the focal lengths \( \frac{f_1}{f_2} \):

\[ \frac{f_1}{f_2} = \frac{R}{\frac{R}{1.4}} = 1.4 \]

So, the ratio of the focal lengths is:

\[ \frac{f_1}{f_2} = \frac{7}{5} \]

### Step 5: Behavior of Lenses in a Liquid with Refractive Index 1.6

Now, we need to analyze how each lens behaves when submerged in a liquid with a refractive index \( n_l = 1.6 \).

#### For Lens with \( n_1 = 1.5 \):

Using the modified formula for the focal length in a medium:

\[ \frac{1}{f_1} = \frac{n_1}{n_l} - 1 \cdot \frac{2}{R} \]

Substituting the values:

\[ \frac{1}{f_1} = \frac{1.5}{1.6} - 1 \cdot \frac{2}{R} = \frac{1.5 - 1.6}{1.6} \cdot \frac{2}{R} = -\frac{0.1}{1.6} \cdot \frac{2}{R} \]

This results in a negative focal length, indicating that the lens behaves like a diverging lens.

#### For Lens with \( n_2 = 1.7 \):

Using the same modified formula:

\[ \frac{1}{f_2} = \frac{n_2}{n_l} - 1 \cdot \frac{2}{R} \]

Substituting the values:

\[ \frac{1}{f_2} = \frac{1.7}{1.6} - 1 \cdot \frac{2}{R} = \frac{1.7 - 1.6}{1.6} \cdot \frac{2}{R} = \frac{0.1}{1.6} \cdot \frac{2}{R} \]

This results in a positive focal length, indicating that the lens behaves like a converging lens.

### Final Summary

1. The ratio of the focal lengths of the two lenses is \( \frac{7}{5} \).
2. The lens made from glass with refractive index 1.5 behaves like a diverging lens when submerged in the liquid.
3. The lens made from glass with refractive index 1.7 behaves like a converging lens when submerged in the liquid.