An equiconvex lens has a power of 5 dioptre. If it is made of glass of refractive index 1.5. then radius of curvature of its each surface will be ?

Options

A. 20 cm
B. 10 cm
C. 40 cm
D. `prop`

Answer

To find the radius of curvature of each surface of an equiconvex lens with a power of 5 diopters and a refractive index of 1.5, we can use the lens maker's formula. Here’s the step-by-step solution:

### Step 1: Understand the Lens Maker's Formula
The lens maker's formula for a lens in air is given by:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For an equiconvex lens, \( R_1 = R \) and \( R_2 = -R \). Thus, the formula simplifies to:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R} + \frac{1}{-R} \right) = (n - 1) \left( \frac{2}{R} \right) \]

### Step 2: Relate Focal Length and Power
The power \( P \) of a lens is related to its focal length \( f \) by the equation:
\[ P = \frac{1}{f} \]
Given that the power \( P = 5 \) diopters, we can find the focal length:
\[ f = \frac{1}{P} = \frac{1}{5} \text{ meters} \]

### Step 3: Substitute Values into the Lens Maker's Formula
Now we can substitute \( f \) and the refractive index \( n \) into the lens maker's formula:
\[ \frac{1}{f} = (n - 1) \left( \frac{2}{R} \right) \]
Substituting \( n = 1.5 \) and \( f = \frac{1}{5} \):
\[ \frac{1}{\frac{1}{5}} = (1.5 - 1) \left( \frac{2}{R} \right) \]
This simplifies to:
\[ 5 = 0.5 \left( \frac{2}{R} \right) \]

### Step 4: Solve for Radius of Curvature \( R \)
Now we can solve for \( R \):
\[ 5 = \frac{1}{R} \]
This implies:
\[ R = \frac{1}{5} \text{ meters} \]

### Step 5: Convert to Centimeters
To convert meters to centimeters:
\[ R = \frac{1}{5} \times 100 = 20 \text{ centimeters} \]

### Final Answer
The radius of curvature of each surface of the lens is \( R = 20 \) centimeters.

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