A plano-convex lens when silvered in the plane side behaves like a concave mirror of focal length 30 cm. However, when silvered on the convex side it behaves like concave mirror of focal length 10 cm. Then the refractive index of its material will be:

Options

A. `3.0`
B. `2.0`
C. `2.5`
D. `1.5`

Answer

To find the refractive index of the plano-convex lens that behaves like a concave mirror when silvered on either side, we can follow these steps:

### Step 1: Understand the behavior of the lens when silvered
When a plano-convex lens is silvered on the plane side, it behaves like a concave mirror with a focal length \( f_1 = -30 \, \text{cm} \) (negative because it is a concave mirror). When silvered on the convex side, it behaves like a concave mirror with a focal length \( f_2 = -10 \, \text{cm} \).

### Step 2: Use the mirror formula for the first case
For the first case (silvered on the plane side), we can use the mirror formula:
\[ \frac{1}{f} = \frac{2}{R} + \frac{1}{f_m} \]
Where:
- \( f \) is the focal length of the mirror (which is \( -30 \, \text{cm} \)),
- \( R \) is the radius of curvature of the lens,
- \( f_m \) is the focal length of the lens when the plane side is silvered (which is infinite for a plane surface).

Since \( f_m = \infty \), the term \( \frac{1}{f_m} \) becomes zero. Thus, we have:
\[ \frac{1}{-30} = \frac{2}{R} \]
From this, we can find \( R \):
\[ R = -60 \, \text{cm} \]

### Step 3: Use the mirror formula for the second case
For the second case (silvered on the convex side), we again use the mirror formula:
\[ \frac{1}{f} = \frac{2}{R} + \frac{1}{f_m} \]
Where:
- \( f \) is now \( -10 \, \text{cm} \),
- \( R \) is the radius of curvature we just found.

Now, since the convex side is silvered, we need to consider the focal length:
\[ \frac{1}{-10} = \frac{2}{R} + \frac{1}{f_m} \]
Here, \( f_m \) is the focal length of the lens when it is not silvered, which we can denote as \( f \).

### Step 4: Substitute \( R \) into the equation
Substituting \( R = -60 \, \text{cm} \) into the equation:
\[ \frac{1}{-10} = \frac{2}{-60} + \frac{1}{f} \]
This simplifies to:
\[ -\frac{1}{10} = -\frac{1}{30} + \frac{1}{f} \]
Rearranging gives:
\[ \frac{1}{f} = -\frac{1}{10} + \frac{1}{30} \]
Finding a common denominator (which is 30):
\[ \frac{1}{f} = -\frac{3}{30} + \frac{1}{30} = -\frac{2}{30} = -\frac{1}{15} \]
Thus, \( f = -15 \, \text{cm} \).

### Step 5: Use the lens maker's formula to find the refractive index
The lens maker's formula is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For a plano-convex lens:
- \( R_1 = R = -60 \, \text{cm} \) (convex side),
- \( R_2 = \infty \) (plane side).

Substituting these values into the lens maker's formula:
\[ \frac{1}{-15} = (\mu - 1) \left( \frac{1}{-60} - 0 \right) \]
This simplifies to:
\[ \frac{1}{-15} = -\frac{\mu - 1}{60} \]
Cross-multiplying gives:
\[ 60 = 15(\mu - 1) \]
Thus:
\[ \mu - 1 = 4 \implies \mu = 5 \]

### Final Answer
The refractive index of the material of the lens is \( \mu = 1.5 \).