The power in dioptre of an equi-convex lens with radii of curvature of 10cm and refractive index 1.6 is

Options

A. `+12`
B. `+18`
C. `+1.2`
D. `+1.8`

Explanation

To find the power of an equi-convex lens with given radii of curvature and refractive index, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of curvature (R) = 10 cm
- Refractive index (μ) = 1.6

2. **Understand the Lens Formula:**
The formula to find the focal length (F) of a lens is given by:
\[ \frac{1}{F} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For an equi-convex lens, \( R_1 = R \) (positive) and \( R_2 = -R \) (negative). Thus:
\[ R_1 = 10 \, \text{cm} \quad \text{and} \quad R_2 = -10 \, \text{cm} \]

3. **Substitute the Values into the Lens Formula:**
\[ \frac{1}{F} = (1.6 - 1) \left( \frac{1}{10} - \frac{1}{-10} \right) \]
This simplifies to:
\[ \frac{1}{F} = 0.6 \left( \frac{1}{10} + \frac{1}{10} \right) \]
\[ \frac{1}{F} = 0.6 \left( \frac{2}{10} \right) = 0.6 \times 0.2 = 0.12 \]

4. **Calculate the Focal Length (F):**
\[ F = \frac{1}{0.12} \, \text{m} \]
Converting this to centimeters:
\[ F = \frac{100}{12} \, \text{cm} \approx 8.33 \, \text{cm} \]

5. **Convert Focal Length to Meters:**
\[ F = \frac{100}{12} \times \frac{1}{100} \, \text{m} = \frac{1}{12} \, \text{m} \]

6. **Calculate the Power (P):**
The power of the lens in diopters (D) is given by:
\[ P = \frac{1}{F} \, \text{(in meters)} \]
\[ P = \frac{1}{\frac{1}{12}} = 12 \, \text{D} \]

### Final Answer:
The power of the equi-convex lens is **12 diopters**.

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