A thin plano-convex lens fits exactly into a thin plano-concave lens. Their plane surfaces are parallel to each other. The lenses are made of different material of refractive indices `mu_(1)` and `mu_(2)` and R is the radius of curvature of the curved surfaces of the lenses, if rays are incident parallel to principal axis on lens of refractive index.

Options

A. `mu_(1)` then the focal length of the combination is `(R)/(mu_(1)-mu_(2))`
B. `mu_(2)` then the focal length of combination is `(2R)/(mu_(2)mu_(1))`
C. `mu_(1)` then the focal length of the combination is `(R)/(2(mu_(1)-mu_(2)))`
D. `mu_(2)` then the focal length of the combination is `(R)/((mu_(2)-mu_(1)))`

Explanation

To solve the problem regarding the equivalent focal length of a thin plano-convex lens and a thin plano-concave lens, we will use the lens maker's formula and the concept of equivalent focal lengths for lenses in combination.

### Step-by-Step Solution:

1. **Identify the Lenses and Their Properties**:
- We have a plano-convex lens with refractive index \( \mu_1 \) and a plano-concave lens with refractive index \( \mu_2 \).
- The radius of curvature for both lenses is \( R \).

2. **Lens Maker's Formula**:
- The lens maker's formula for a lens is given by:
\[ \frac{1}{F} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
- For the plano-convex lens:
- \( R_1 = R \) (convex surface)
- \( R_2 = \infty \) (plane surface)
- Therefore, the focal length \( F_1 \) is:
\[ \frac{1}{F_1} = (\mu_1 - 1) \left( \frac{1}{R} - 0 \right) = \frac{\mu_1 - 1}{R} \]
\[ F_1 = \frac{R}{\mu_1 - 1} \]

3. **For the Plano-Concave Lens**:
- For the plano-concave lens:
- \( R_1 = -R \) (concave surface)
- \( R_2 = \infty \) (plane surface)
- Therefore, the focal length \( F_2 \) is:
\[ \frac{1}{F_2} = (\mu_2 - 1) \left( \frac{-1}{R} - 0 \right) = -\frac{\mu_2 - 1}{R} \]
\[ F_2 = -\frac{R}{\mu_2 - 1} \]

4. **Calculate the Equivalent Focal Length**:
- When the two lenses are combined, the equivalent focal length \( F_{eq} \) is given by:
\[ \frac{1}{F_{eq}} = \frac{1}{F_1} + \frac{1}{F_2} \]
- Substituting the values of \( F_1 \) and \( F_2 \):
\[ \frac{1}{F_{eq}} = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R} \]
\[ \frac{1}{F_{eq}} = \frac{(\mu_1 - 1) - (\mu_2 - 1)}{R} = \frac{\mu_1 - \mu_2}{R} \]
- Therefore, the equivalent focal length is:
\[ F_{eq} = \frac{R}{\mu_1 - \mu_2} \]

5. **Direction of Focal Length**:
- If \( \mu_1 > \mu_2 \), the focal length \( F_{eq} \) is positive, indicating a converging lens.
- If \( \mu_1 < \mu_2 \), the focal length \( F_{eq} \) is negative, indicating a diverging lens.