Consider an equiconvex lens of radius of curvature R and focal length f. If `fgtR` , the refractive index `mu` of the material of the lens

Options

A. is greater than zero but less than 1.5
B. is greater than 1.5 but less then 2.0
C. is greater than 1.0 but less than 1.5
D. None of these

Explanation

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

### Step 1: Understand the Lens Maker's Formula
The lens maker's formula for a lens is given by:

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

For an equiconvex lens, both radii of curvature are equal in magnitude but opposite in sign. Thus, we can denote:

- \( R_1 = R \) (for the first surface)
- \( R_2 = -R \) (for the second surface)

### Step 2: Substitute the Values into the Formula
Substituting the values of \( R_1 \) and \( R_2 \) into the lens maker's formula:

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

This simplifies to:

\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R} + \frac{1}{R} \right) = (\mu - 1) \left( \frac{2}{R} \right) \]

### Step 3: Rearranging the Equation
Rearranging the equation gives:

\[ \frac{1}{f} = \frac{2(\mu - 1)}{R} \]

Multiplying both sides by \( R \):

\[ R \cdot \frac{1}{f} = 2(\mu - 1) \]

### Step 4: Solve for \( \mu \)
Now, solving for \( \mu \):

\[ \mu - 1 = \frac{R}{2f} \]

Thus,

\[ \mu = 1 + \frac{R}{2f} \]

### Step 5: Analyze the Condition Given
We are given that \( f > R \). This implies:

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

Consequently,

\[ \frac{R}{2f} < \frac{1}{2} \]

### Step 6: Substitute Back into the Equation for \( \mu \)
From the previous step, we can conclude:

\[ \mu < 1 + \frac{1}{2} = 1.5 \]

### Step 7: Determine the Range of \( \mu \)
Since \( f \) is positive and \( R \) is also positive, we know that \( \mu \) must be greater than 1 (as it is a refractive index). Thus, we have:

\[ 1 < \mu < 1.5 \]

### Conclusion
The refractive index \( \mu \) of the material of the lens is greater than 1 but less than 1.5. Therefore, the correct option is:

**Option C: \( \mu \) is greater than 1 but less than 1.5.**

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