What is the refractive index of material of a plano-convex lens , if the radius of curvature of the convex surface is 10 cm and focal length of the lens is 30 cm ?

Options

A. `6/5`
B. `7/4`
C. `2/3`
D. `4/3`

Explanation

To find the refractive index (μ) of a plano-convex lens given the radius of curvature (R) of the convex surface and the focal length (f) of the lens, we can use the lens maker's formula:

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

### Step 1: Identify the values
- The focal length of the lens, \( f = 30 \) cm.
- The radius of curvature of the convex surface, \( R_2 = 10 \) cm (since it is convex, we take it as positive).
- The plano surface has an infinite radius of curvature, \( R_1 = \infty \).

### Step 2: Substitute the values into the lens maker's formula
Using the lens maker's formula:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting \( R_1 \) and \( R_2 \):
\[ \frac{1}{30} = (\mu - 1) \left( \frac{1}{\infty} - \frac{1}{10} \right) \]

### Step 3: Simplify the equation
Since \( \frac{1}{\infty} \) is negligible (approaches 0), the equation simplifies to:
\[ \frac{1}{30} = (\mu - 1) \left( 0 - \frac{1}{10} \right) \]
This can be rewritten as:
\[ \frac{1}{30} = -(\mu - 1) \frac{1}{10} \]

### Step 4: Remove the negative sign and solve for \( \mu - 1 \)
Multiplying both sides by -10 gives:
\[ -\frac{10}{30} = \mu - 1 \]
This simplifies to:
\[ -\frac{1}{3} = \mu - 1 \]

### Step 5: Solve for \( \mu \)
Adding 1 to both sides:
\[ \mu = 1 - \frac{1}{3} = \frac{2}{3} \]

### Step 6: Correct the calculation
It seems there was a mistake in the sign. The correct equation should be:
\[ \frac{1}{30} = (\mu - 1) \left( \frac{1}{10} \right) \]
Thus:
\[ \mu - 1 = \frac{1}{30} \times 10 = \frac{1}{3} \]
Adding 1 gives:
\[ \mu = 1 + \frac{1}{3} = \frac{4}{3} \]

### Final Answer
The refractive index of the material of the plano-convex lens is:
\[ \mu = \frac{4}{3} \]

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