The radius of curvature of the convex face of a plano convex lens is 12cm and the refractive index of the material of the lends 1.5. Then, the focal length of the lens is

Options

A. 6cm
B. 12cm
C. 18cm
D. 24cm

Explanation

To find the focal length of a plano-convex lens, we can use the lens maker's formula:

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

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 (convex surface),
- \( R_2 \) is the radius of curvature of the second surface (plano surface).

### Step 1: Identify the values
- Given:
- Radius of curvature of the convex face, \( R_1 = +12 \, \text{cm} \) (positive because it is convex)
- The second surface is plano, so \( R_2 = \infty \) (the radius of curvature for a flat surface is considered infinite)
- Refractive index, \( \mu = 1.5 \)

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

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

Substituting the known values:

\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{12} - \frac{1}{\infty} \right) \]

### Step 3: Simplify the equation
Since \( \frac{1}{\infty} = 0 \), the equation simplifies to:

\[ \frac{1}{f} = 0.5 \left( \frac{1}{12} - 0 \right) \]

\[ \frac{1}{f} = 0.5 \cdot \frac{1}{12} \]

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

\[ \frac{1}{f} = \frac{0.5}{12} = \frac{1}{24} \]

### Step 5: Find the focal length \( f \)
Taking the reciprocal gives us:

\[ f = 24 \, \text{cm} \]

### Conclusion
The focal length of the plano-convex lens is \( 24 \, \text{cm} \).

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