The refractive index of a material of a plano concave lens is `5/3`. Its radius of curvature is 0.3 m. Focal length of the lens in air is

Options

A. 0.45 m
B. -0.6m
C. `0.7m`
D. `1m`

Explanation

To find the focal length of a plano-concave lens in air, we can use the lens maker's formula. The formula for a lens is given by:

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

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 1: Identify the values
Given:
- Refractive index, \( \mu = \frac{5}{3} \)
- Radius of curvature, \( R = 0.3 \, \text{m} \)

For a plano-concave lens:
- The first surface is flat, so \( R_1 = \infty \) (flat surface),
- The second surface is concave, so \( R_2 = -0.3 \, \text{m} \) (negative because it is concave).

### Step 2: Substitute the values into the formula
Using the lens maker's formula:

\[ \frac{1}{f} = \left( \frac{5}{3} - 1 \right) \left( \frac{1}{\infty} - \frac{1}{-0.3} \right) \]

### Step 3: Simplify the equation
Calculating \( \mu - 1 \):

\[ \frac{5}{3} - 1 = \frac{5}{3} - \frac{3}{3} = \frac{2}{3} \]

Now, substituting into the formula:

\[ \frac{1}{f} = \left( \frac{2}{3} \right) \left( 0 - \left( -\frac{1}{0.3} \right) \right) \]

This simplifies to:

\[ \frac{1}{f} = \left( \frac{2}{3} \right) \left( \frac{1}{0.3} \right) \]

### Step 4: Calculate \( \frac{1}{0.3} \)
Calculating \( \frac{1}{0.3} \):

\[ \frac{1}{0.3} = \frac{10}{3} \]

### Step 5: Substitute back into the equation
Now substituting this back:

\[ \frac{1}{f} = \left( \frac{2}{3} \right) \left( \frac{10}{3} \right) = \frac{20}{9} \]

### Step 6: Find \( f \)
Now, taking the reciprocal to find \( f \):

\[ f = \frac{9}{20} \, \text{m} = 0.45 \, \text{m} \]

### Final Answer
The focal length of the plano-concave lens in air is \( 0.45 \, \text{m} \).

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