The focal length of equiconvex lens of retractive index, `mu=1.5` and radius of curvature, R is

Options

A. `R / 1.5`
B. `1.5/R`
C. R
D. 1.5R

Answer

To find the focal length of an equiconvex lens with a refractive index (μ) of 1.5 and radius of curvature (R), we will use the lens maker's formula. Here’s a step-by-step solution:

### Step 1: Write down the lens maker's formula
The lens maker's formula is given by:

\[ \frac{1}{f} = \left(\frac{\mu_1}{\mu_2} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \]

### Step 2: Identify the parameters
For an equiconvex lens:
- The refractive index of the lens (μ1) = 1.5
- The refractive index of air (μ2) = 1 (approximately)
- For an equiconvex lens, the radius of curvature for the first surface (R1) = R and for the second surface (R2) = -R (the negative sign indicates that the second surface is concave relative to the incoming light).

### Step 3: Substitute the values into the formula
Substituting the values into the lens maker's formula:

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

### Step 4: Simplify the equation
Now simplify the equation:

1. Calculate \(\frac{1.5}{1} - 1\):

\[ \frac{1.5}{1} - 1 = 0.5 \]

2. Calculate \(\frac{1}{R} - \frac{1}{-R}\):

\[ \frac{1}{R} - \frac{1}{-R} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \]

3. Substitute these results back into the equation:

\[ \frac{1}{f} = 0.5 \cdot \frac{2}{R} \]

### Step 5: Further simplify
Now simplify the right side:

\[ \frac{1}{f} = \frac{1}{R} \]

### Step 6: Solve for the focal length (f)
Taking the reciprocal gives us:

\[ f = R \]

### Final Answer
Thus, the focal length of the equiconvex lens in terms of the radius of curvature R is:

\[ f = R \]

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