A plano-concave lens is made of glass of refractive index 1.5 and the radius of curvature of its curved face is 100 cm. What is the power of the lens?

Options

A. `+0.5D`
B. `-0.5D`
C. `-2D`
D. `+2D`

Explanation

To find the power of a plano-concave lens made of glass with a refractive index of 1.5 and a radius of curvature of 100 cm, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the parameters:**
- Refractive index of the lens, \( \mu = 1.5 \)
- Radius of curvature of the curved face, \( R = 100 \, \text{cm} \)
- For a plano-concave lens, the plano side can be considered as having an infinite radius of curvature, \( R_1 = \infty \).

2. **Use the Lensmaker's Formula:**
The Lensmaker's formula for a lens is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Where:
- \( f \) is the focal length,
- \( R_1 \) is the radius of curvature of the first surface,
- \( R_2 \) is the radius of curvature of the second surface.

3. **Substituting the values:**
Since \( R_1 = \infty \) and \( R_2 = -100 \, \text{cm} \) (the negative sign indicates that the surface is concave):
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{\infty} - \frac{1}{-100} \right) \]
This simplifies to:
\[ \frac{1}{f} = 0.5 \left( 0 + \frac{1}{100} \right) \]
\[ \frac{1}{f} = 0.5 \times \frac{1}{100} \]
\[ \frac{1}{f} = \frac{0.5}{100} = \frac{1}{200} \]

4. **Finding the focal length:**
Taking the reciprocal gives us the focal length:
\[ f = 200 \, \text{cm} = 2 \, \text{m} \]

5. **Calculating the power of the lens:**
The power \( P \) of a lens is given by:
\[ P = \frac{1}{f} \quad \text{(in meters)} \]
Therefore:
\[ P = \frac{1}{-2} = -0.5 \, \text{D} \]
(The negative sign indicates that it is a diverging lens.)

### Final Answer:
The power of the plano-concave lens is \( -0.5 \, \text{D} \).

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