A spherical convex surface separates object and image space of refractive index `1.0 and 4/3.` If radius of curvature of the surface is `10cm,` find its power.

Explanation

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

### Step 1: Understand the given values
- Refractive index of the object space (air) \( \mu_1 = 1.0 \)
- Refractive index of the image space \( \mu_2 = \frac{4}{3} \)
- Radius of curvature \( R = 10 \, \text{cm} \)

### Step 2: Identify the sign convention
Since the surface is convex, we will take the radius of curvature as positive:
- \( R = +10 \, \text{cm} \)

### Step 3: Use the refraction formula
We will use the formula for refraction at a spherical surface:
\[ \frac{\mu_2}{V} - \frac{\mu_1}{U} = \frac{\mu_2 - \mu_1}{R} \]
Where:
- \( V \) is the image distance
- \( U \) is the object distance (for parallel rays, \( U = -\infty \))

### Step 4: Substitute the known values
Substituting the values into the formula:
\[ \frac{\frac{4}{3}}{V} - \frac{1}{-\infty} = \frac{\frac{4}{3} - 1}{10} \]
Since \( \frac{1}{-\infty} \) approaches 0, we can simplify:
\[ \frac{\frac{4}{3}}{V} = \frac{\frac{4}{3} - 1}{10} \]
Calculating \( \frac{4}{3} - 1 \):
\[ \frac{4}{3} - 1 = \frac{4}{3} - \frac{3}{3} = \frac{1}{3} \]
Thus, we have:
\[ \frac{\frac{4}{3}}{V} = \frac{\frac{1}{3}}{10} \]

### Step 5: Solve for \( V \)
Cross-multiplying gives:
\[ \frac{4}{3} \cdot 10 = \frac{1}{3} \cdot V \]
\[ \frac{40}{3} = \frac{1}{3} V \]
Multiplying both sides by 3:
\[ 40 = V \]
So, the image distance \( V = 40 \, \text{cm} \).

### Step 6: Calculate the focal length
For a convex surface, the focal length \( f \) is equal to the image distance when the object is at infinity:
\[ f = V = 40 \, \text{cm} \]

### Step 7: Convert focal length to meters
Convert \( f \) to meters:
\[ f = 40 \, \text{cm} = 0.4 \, \text{m} \]

### Step 8: Calculate the power of the lens
The power \( P \) of a lens is given by:
\[ P = \frac{1}{f} \quad (\text{in meters}) \]
Substituting the value of \( f \):
\[ P = \frac{1}{0.4} = 2.5 \, \text{diopters} \]

### Final Answer
The power of the lens is \( 2.5 \, \text{D} \).

---