Show that the focal length of a thin lens is not changed when the lens is rotated so that the left and the right surfaces are interchanged.

Explanation

To show that the focal length of a thin lens does not change when the lens is rotated such that the left and right surfaces are interchanged, we can follow these steps:

### Step-by-Step Solution

1. **Define the Lens and Surfaces**:
Consider a thin lens with two surfaces. Let the first surface be denoted as \( R_1 \) and the second surface as \( R_2 \). The lens has a refractive index \( \mu \).

2. **Apply the Lensmaker's Formula**:
The Lensmaker's formula relates the focal length \( f \) of a lens to its radii of curvature and the refractive index. For the original orientation of the lens, the formula is given by:
\[ \frac{1}{f_1} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Here, \( f_1 \) is the focal length when light comes from the side of \( R_1 \).

3. **Interchange the Surfaces**:
When the lens is rotated, the surfaces are interchanged. Now, the first surface becomes \( R_2 \) and the second surface becomes \( R_1 \). The new focal length \( f_2 \) can be expressed as:
\[ \frac{1}{f_2} = (\mu - 1) \left( \frac{1}{R_2} - \frac{1}{R_1} \right) \]

4. **Simplify the New Expression**:
Notice that the expression for \( f_2 \) can be rewritten:
\[ \frac{1}{f_2} = (\mu - 1) \left( \frac{1}{R_2} - \frac{1}{R_1} \right) = (\mu - 1) \left( -\left( \frac{1}{R_1} - \frac{1}{R_2} \right) \right) \]
This shows that:
\[ \frac{1}{f_2} = -(\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

5. **Relate \( f_1 \) and \( f_2 \)**:
From the original expression for \( f_1 \), we have:
\[ \frac{1}{f_1} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Therefore, we can see that:
\[ \frac{1}{f_2} = -\frac{1}{f_1} \]
This implies that \( f_1 = f_2 \).

6. **Conclusion**:
Since \( f_1 = f_2 \), we conclude that the focal length of the thin lens does not change when the lens is rotated such that the left and right surfaces are interchanged.