Radius of curvature of a equi convex lens is R. Find focal length (`n = 3/2`)

Options

A. `R/2`
B. `-R/2`
C. R
D. -R

Answer

To find the focal length of an equi-convex lens with a radius of curvature \( R \) and a refractive index \( n = \frac{3}{2} \), we can use the lensmaker's formula:

\[ \frac{1}{f} = (n - 1) \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]

### Step-by-step Solution:

1. **Identify the parameters**:
- For an equi-convex lens, both surfaces have the same radius of curvature \( R \).
- Let \( r_1 = R \) (for the first surface) and \( r_2 = -R \) (for the second surface, which is conventionally taken as negative).

2. **Substitute the values into the lensmaker's formula**:
\[ \frac{1}{f} = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{R} - \frac{1}{-R} \right) \]

3. **Simplify the equation**:
- Calculate \( n - 1 \):
\[ n - 1 = \frac{3}{2} - 1 = \frac{1}{2} \]
- Calculate \( \frac{1}{R} - \frac{1}{-R} \):
\[ \frac{1}{R} - \frac{1}{-R} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \]

4. **Combine the results**:
\[ \frac{1}{f} = \frac{1}{2} \cdot \frac{2}{R} \]
\[ \frac{1}{f} = \frac{1}{R} \]

5. **Find the focal length**:
- Taking the reciprocal gives:
\[ f = R \]

### Final Result:
The focal length \( f \) of the equi-convex lens is \( R \).

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