The focal length of equiconvex lens of retractive index, `mu=1.5` and radius of curvature, R is

Options

A. `R / 1.5`
B. `1.5/R`
C. R
D. 1.5R

Answer

To find the focal length of an equiconvex lens with a refractive index (μ) of 1.5 and radius of curvature (R), we can use the lens maker's formula:

\[ \frac{1}{f} = (μ - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step-by-Step Solution:

1. **Identify the Parameters**:
- For an equiconvex lens, both radii of curvature are equal. Let \( R_1 = R \) (for the first surface) and \( R_2 = -R \) (for the second surface, since it is concave with respect to the incoming light).
- The refractive index \( μ = 1.5 \).

2. **Substitute Values into the Lens Maker's Formula**:
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) \]

3. **Simplify the Equation**:
\[ \frac{1}{f} = 0.5 \left( \frac{1}{R} + \frac{1}{R} \right) \]
\[ \frac{1}{f} = 0.5 \left( \frac{2}{R} \right) \]
\[ \frac{1}{f} = \frac{1}{R} \]

4. **Find the Focal Length**:
\[ f = R \]

Thus, the focal length of the equiconvex lens is \( f = R \).

### Final Answer:
The focal length of the equiconvex lens is equal to the radius of curvature \( R \).

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