A double convex lens whose radii of curvature are `R_(1)` and `R_(2)` forms the image of a point object, placed on its axis. If the lens is revered, face to face, then the ratio of the distances of the images in the first and second case will be

Options

A. `(R_(1))/(R_(2))`
B. `(R_(2))/(R_(1))`
C. `1:1`
D. `2:1`

Explanation

To solve the problem, we need to analyze the behavior of a double convex lens when it is placed in two different orientations. We will derive the focal lengths and the corresponding image distances in both cases.

### Step-by-Step Solution:

1. **Understanding the Lens Formula**:
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance. For a double convex lens, the focal length can be calculated using the radii of curvature.

2. **Focal Length Calculation**:
The focal length \( f \) of a double convex lens is given by:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where \( n \) is the refractive index of the lens material, and \( R_1 \) and \( R_2 \) are the radii of curvature of the two surfaces of the lens.

3. **First Case (Normal Orientation)**:
- When the lens is in its normal orientation, we denote the focal length as \( f_1 \).
- The image distance \( v_1 \) can be calculated using the lens formula, with the object distance \( u \) being negative (as per sign convention):
\[ \frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{(-u)} \]
Rearranging gives:
\[ v_1 = \frac{fu}{f + u} \]

4. **Second Case (Reversed Orientation)**:
- When the lens is reversed, the radii of curvature change places, so we denote the new focal length as \( f_2 \):
\[ \frac{1}{f_2} = (n - 1) \left( \frac{1}{R_2} - \frac{1}{R_1} \right) \]
- The image distance \( v_2 \) can be calculated similarly:
\[ v_2 = \frac{f_2u}{f_2 + u} \]

5. **Comparing the Two Cases**:
- Since the focal lengths \( f_1 \) and \( f_2 \) are related by the same formula but with \( R_1 \) and \( R_2 \) swapped, we find that:
\[ f_1 = f_2 \]
- Therefore, the image distances \( v_1 \) and \( v_2 \) will also be equal, leading to:
\[ \frac{v_1}{v_2} = 1 \]

6. **Conclusion**:
The ratio of the distances of the images in the first and second case is:
\[ \text{Ratio} = 1:1 \]

### Final Answer:
The ratio of the distances of the images in the first and second case is \( 1:1 \).