A plano-convex lens has refractive index 1.6 and radius of curvature 60 cm. What is the focal length of the lens?

Options

A. 50 cm
B. 100 cm
C. 200 cm
D. 150 cm

Explanation

To find the focal length of a plano-convex lens with a given refractive index and radius of curvature, we can use the lens maker's formula. Here’s a step-by-step solution:

### Step 1: Identify the parameters
- Refractive index (μ) = 1.6
- Radius of curvature (R) = 60 cm
- For a plano-convex lens, one surface is flat (plane) and the other is convex. Thus:
- R1 (for the flat surface) = ∞ (infinity)
- R2 (for the convex surface) = -60 cm (according to the sign convention, the radius of curvature for a convex surface is taken as negative)

### Step 2: Write the lens maker's formula
The lens maker's formula is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step 3: Substitute the values into the formula
Substituting the values we have:
- μ = 1.6
- R1 = ∞
- R2 = -60 cm

So the equation becomes:
\[ \frac{1}{f} = (1.6 - 1) \left( \frac{1}{\infty} - \frac{1}{-60} \right) \]

### Step 4: Simplify the equation
Calculating the terms:
- \( \mu - 1 = 0.6 \)
- \( \frac{1}{\infty} = 0 \)
- \( -\frac{1}{-60} = \frac{1}{60} \)

Now substituting these values:
\[ \frac{1}{f} = 0.6 \left( 0 + \frac{1}{60} \right) \]
\[ \frac{1}{f} = 0.6 \cdot \frac{1}{60} \]

### Step 5: Calculate \( \frac{1}{f} \)
Calculating \( 0.6 \cdot \frac{1}{60} \):
\[ \frac{1}{f} = \frac{0.6}{60} = \frac{6}{600} = \frac{1}{100} \]

### Step 6: Find the focal length \( f \)
Taking the reciprocal gives:
\[ f = 100 \text{ cm} \]

### Final Answer
The focal length of the plano-convex lens is **100 cm**.

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