The focal length of a plano-convex lens is `f` and its refractive index is 1.5 it is kept over a plane galss plate with its curved surface touching the glass plate. The gap between the lens and the glass plate is filled by a liquid. As a result the effective focal length of the combination becomes `2f`. Then the refractive index of the liquid is

Options

A. 1.5
B. 2
C. 1.25
D. 1.33

Explanation

To find the refractive index of the liquid that fills the gap between the plano-convex lens and the glass plate, we can follow these steps:

### Step 1: Understand the Setup
We have a plano-convex lens with focal length \( f \) and refractive index \( \mu = 1.5 \). The lens is placed on a plane glass plate, and the gap between the lens and the glass plate is filled with a liquid. The effective focal length of the combination becomes \( 2f \).

### Step 2: Write the Formula for Effective Focal Length
The formula for the effective focal length \( F \) of two lenses in contact is given by:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
where \( f_1 \) is the focal length of the plano-convex lens and \( f_2 \) is the focal length of the liquid layer acting as a lens.

### Step 3: Identify the Focal Lengths
For the plano-convex lens:
- \( f_1 = f \)

For the liquid layer, which behaves as a plano-concave lens:
- Let the focal length of the liquid be \( f_2 \).

The effective focal length of the combination is given as \( 2f \):
\[ \frac{1}{2f} = \frac{1}{f} + \frac{1}{f_2} \]

### Step 4: Rearranging the Equation
Rearranging the above equation gives:
\[ \frac{1}{f_2} = \frac{1}{2f} - \frac{1}{f} \]
Finding a common denominator:
\[ \frac{1}{f_2} = \frac{1 - 2}{2f} = -\frac{1}{2f} \]
Thus,
\[ f_2 = -2f \]

### Step 5: Use the Lensmaker's Formula for the Liquid
The lensmaker's formula is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For the liquid layer:
- The radius of curvature \( R_1 = -r \) (for the curved surface of the lens),
- The radius of curvature \( R_2 = \infty \) (for the flat surface of the glass plate).

Thus, the formula for the liquid becomes:
\[ \frac{1}{f_2} = (\mu_{liquid} - 1) \left( \frac{1}{-r} - 0 \right) \]
Substituting \( f_2 = -2f \):
\[ -\frac{1}{2f} = (\mu_{liquid} - 1) \left( -\frac{1}{r} \right) \]

### Step 6: Relate the Focal Lengths
Now, for the plano-convex lens:
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{r} - 0 \right) = \frac{0.5}{r} \]
Thus, \( r = 0.5f \).

### Step 7: Substitute \( r \) into the Liquid Equation
Substituting \( r \) into the equation for the liquid:
\[ -\frac{1}{2f} = (\mu_{liquid} - 1) \left( -\frac{1}{0.5f} \right) \]
This simplifies to:
\[ -\frac{1}{2f} = -2(\mu_{liquid} - 1) \frac{1}{f} \]
Cancelling \( -\frac{1}{f} \) from both sides gives:
\[ \frac{1}{2} = 2(\mu_{liquid} - 1) \]
Thus,
\[ \mu_{liquid} - 1 = \frac{1}{4} \]
So,
\[ \mu_{liquid} = 1 + \frac{1}{4} = 1.25 \]

### Conclusion
The refractive index of the liquid is \( \mu_{liquid} = 1.25 \).