The radius of curvature of the convex surface of a thin plano-convex lens is 15 cm and the refractive index of its material is 1.6. The power of the lens is

Options

A. `+1D`
B. `-2D`
C. `+3D`
D. `+4D`

Explanation

To find the power of the plano-convex lens, we will use the lens maker's formula and follow these steps:

### Step 1: Understand the Lens Maker's Formula
The lens maker's formula is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where:
- \( f \) is the focal length of the lens,
- \( \mu \) is the refractive index of the lens material,
- \( R_1 \) is the radius of curvature of the first surface,
- \( R_2 \) is the radius of curvature of the second surface.

### Step 2: Identify the Values
For a plano-convex lens:
- The convex surface (R1) has a radius of curvature \( R_1 = +15 \, \text{cm} \) (positive because it's convex).
- The plane surface (R2) has an infinite radius of curvature \( R_2 = \infty \) (the plane surface does not bend light).
- The refractive index \( \mu = 1.6 \).

### Step 3: Substitute Values into the Formula
Substituting the values into the lens maker's formula:
\[ \frac{1}{f} = (1.6 - 1) \left( \frac{1}{15} - \frac{1}{\infty} \right) \]
Since \( \frac{1}{\infty} = 0 \), the equation simplifies to:
\[ \frac{1}{f} = 0.6 \left( \frac{1}{15} \right) \]

### Step 4: Calculate the Focal Length
Calculating the right side:
\[ \frac{1}{f} = 0.6 \times \frac{1}{15} = \frac{0.6}{15} = \frac{0.6}{15} = \frac{6}{150} = \frac{1}{25} \]
Thus, the focal length \( f \) is:
\[ f = 25 \, \text{cm} \]

### Step 5: Calculate the Power of the Lens
The power \( P \) of the lens is given by:
\[ P = \frac{100}{f \, (\text{in cm})} \]
Substituting the focal length:
\[ P = \frac{100}{25} = 4 \, \text{diopters} \]

### Final Answer
The power of the lens is \( +4 \, \text{diopters} \).

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