A plano convex lens fits exactly into a plano concave lens. Their plane surfaces are parallel to each other. If the lenses are made of different materials refractive indices `mu_(1)` and `mu_(2)` and R is the radius curvature of the curved surface of the lenses, the focal length of the combination is

Options

A. `(R)/(mu_(1)-mu_(2))`
B. `(2R)/(mu_(2)-mu_(1))`
C. `(R)/(2(mu_(1)-mu_(2)))`
D. `(R)/(2-(mu_(1)+mu_(2)))`

Answer

To find the focal length of the combination of a plano-convex lens and a plano-concave lens, we can use the lens maker's formula for each lens and then combine their focal lengths.

### Step-by-Step Solution:

1. **Understanding the Lenses**:
- We have a plano-convex lens and a plano-concave lens. The plano-convex lens has a convex surface facing outward and a flat surface, while the plano-concave lens has a concave surface facing inward and a flat surface.

2. **Lens Maker's Formula**:
- The lens maker's formula is given by:
\[ \frac{1}{f} = \mu - 1 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
- Here, \(f\) is the focal length, \(\mu\) 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. **Focal Length of the Plano-Convex Lens**:
- For the plano-convex lens:
- \(R_1 = \infty\) (flat surface)
- \(R_2 = -R\) (curved surface)
- Applying the lens maker's formula:
\[ \frac{1}{f_1} = \mu_1 - 1 \left( \frac{1}{\infty} - \frac{1}{-R} \right) = \mu_1 - 1 \left( 0 + \frac{1}{R} \right) = \frac{\mu_1 - 1}{R} \]
- Thus, the focal length \(f_1\) of the plano-convex lens is:
\[ f_1 = \frac{R}{\mu_1 - 1} \]

4. **Focal Length of the Plano-Concave Lens**:
- For the plano-concave lens:
- \(R_1 = -R\) (curved surface)
- \(R_2 = \infty\) (flat surface)
- Applying the lens maker's formula:
\[ \frac{1}{f_2} = \mu_2 - 1 \left( \frac{1}{-R} - \frac{1}{\infty} \right) = \mu_2 - 1 \left( -\frac{1}{R} - 0 \right) = -\frac{\mu_2 - 1}{R} \]
- Thus, the focal length \(f_2\) of the plano-concave lens is:
\[ f_2 = -\frac{R}{\mu_2 - 1} \]

5. **Focal Length of the Combination**:
- The total focal length \(F\) of the combination of the two lenses is given by:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
- Substituting the values of \(f_1\) and \(f_2\):
\[ \frac{1}{F} = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R} \]
- Simplifying this:
\[ \frac{1}{F} = \frac{\mu_1 - 1 - (\mu_2 - 1)}{R} = \frac{\mu_1 - \mu_2}{R} \]
- Therefore, the focal length \(F\) of the combination is:
\[ F = \frac{R}{\mu_1 - \mu_2} \]

### Final Answer:
The focal length of the combination is:
\[ F = \frac{R}{\mu_1 - \mu_2} \]