An equiconvex lens of glass of focal length 0.1 metre is cut along a plane perpendicular to principle axis into two equal parts. The ratio of focal length of new lenses formed is:

Options

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

Answer

To solve the problem, we need to analyze the situation of cutting an equiconvex lens into two equal parts and determining the focal lengths of the resulting lenses.

### Step-by-Step Solution:

1. **Understanding the Original Lens**:
- We start with an equiconvex lens with a focal length \( F = 0.1 \) meters.
- An equiconvex lens has two surfaces with equal radius of curvature, denoted as \( R \).

2. **Cutting the Lens**:
- When we cut the lens along a plane perpendicular to the principal axis, we create two new lenses.
- Each new lens will have one flat surface and one curved surface.

3. **Identifying the Surfaces**:
- For the first new lens (let's call it Lens 1), the curved surface has a radius of curvature \( R \) and the flat surface has an infinite radius of curvature (\( R_1 = R \), \( R_2 = \infty \)).
- For the second new lens (Lens 2), the situation is reversed: the flat surface has an infinite radius of curvature and the curved surface has a radius of curvature of \( R \) (i.e., \( R_1 = \infty \), \( R_2 = -R \)).

4. **Using the Lens Maker's Formula**:
- The lens maker's formula is given by:
\[ \frac{1}{F} = (μ - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
- For Lens 1:
\[ \frac{1}{F_1} = (μ - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = (μ - 1) \left( \frac{1}{R} \right) \]
\[ F_1 = \frac{R}{μ - 1} \]

- For Lens 2:
\[ \frac{1}{F_2} = (μ - 1) \left( \frac{1}{\infty} - \left(-\frac{1}{R}\right) \right) = (μ - 1) \left( \frac{1}{R} \right) \]
\[ F_2 = \frac{R}{μ - 1} \]

5. **Comparing the Focal Lengths**:
- From the calculations, we find that:
\[ F_1 = F_2 \]
- Therefore, the ratio of the focal lengths of the two new lenses is:
\[ \frac{F_1}{F_2} = 1 \]

6. **Final Answer**:
- The ratio of the focal lengths of the new lenses formed is \( 1:1 \).