An equiconvex lens is made from glass of refractive index 1.5. If the radius of each surface is changed from 5cm to 6cm, then the power

Options

A. remains unchanged
B. increases by 3.33D
C. decreases by 3.33D
D. decreases by 5.5D

Explanation

To solve the problem step by step, we will use the lens maker's formula and the concept of power of a lens.

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

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

where:
- \( f \) is the focal length of the lens,
- \( n \) 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 2: Identify the parameters
From the problem:
- The refractive index \( n = 1.5 \).
- The initial radius of curvature \( R_1 = 5 \, \text{cm} \) and \( R_2 = -5 \, \text{cm} \) (since it is an equiconvex lens, the second radius is negative).
- The new radius of curvature \( R_1' = 6 \, \text{cm} \) and \( R_2' = -6 \, \text{cm} \).

### Step 3: Calculate the initial focal length
Using the initial radii:

\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{5} - \frac{1}{-5} \right) \]

Calculating the right side:

\[ = 0.5 \left( \frac{1}{5} + \frac{1}{5} \right) = 0.5 \left( \frac{2}{5} \right) = \frac{1}{5} \]

Thus, the initial focal length \( f \) is:

\[ f = 5 \, \text{cm} \]

### Step 4: Calculate the initial power
The power \( P \) of a lens is given by:

\[ P = \frac{1}{f} \text{ (in meters)} \]

Converting \( f \) to meters:

\[ f = 0.05 \, \text{m} \]

So,

\[ P = \frac{1}{0.05} = 20 \, \text{diopters} \]

### Step 5: Calculate the new focal length
Now, using the new radii:

\[ \frac{1}{f'} = (1.5 - 1) \left( \frac{1}{6} - \frac{1}{-6} \right) \]

Calculating the right side:

\[ = 0.5 \left( \frac{1}{6} + \frac{1}{6} \right) = 0.5 \left( \frac{2}{6} \right) = \frac{1}{6} \]

Thus, the new focal length \( f' \) is:

\[ f' = 6 \, \text{cm} \]

### Step 6: Calculate the new power
Again, converting \( f' \) to meters:

\[ f' = 0.06 \, \text{m} \]

So,

\[ P' = \frac{1}{0.06} \approx 16.67 \, \text{diopters} \]

### Step 7: Calculate the change in power
The change in power \( \Delta P \) is:

\[ \Delta P = P - P' = 20 - 16.67 \approx 3.33 \, \text{diopters} \]

### Conclusion
The power of the lens decreases by approximately \( 3.33 \, \text{diopters} \).

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