A thin liquid convex lens is formed in glass. Refractive index of liquid is 4/3 and that of glass is 3/2. If f is the focal length of the liquid lens is air, its focal length and nature in the glass is

Options

A. f, convex
B. f, concave
C. 2f, concave
D. 3f, concave

Answer

To solve the problem step by step, we will use the lens maker's formula and the given refractive indices to find the focal length of the liquid lens when placed in glass.

### Step 1: Understand the lens maker's formula
The lens maker's formula is given by:

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

Where:
- \( f \) is the focal length of the lens,
- \( \mu \) is the refractive index of the lens material relative to the surrounding medium,
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

### Step 2: Calculate the focal length of the liquid lens in air
Given:
- Refractive index of liquid, \( \mu_l = \frac{4}{3} \)
- Refractive index of air, \( \mu_a = 1 \)

Using the lens maker's formula for the lens in air:

\[ \frac{1}{f} = \left( \frac{\frac{4}{3} - 1}{R_1} - \frac{\frac{4}{3} - 1}{R_2} \right) \]

Since \( R_1 \) is positive and \( R_2 \) is negative, we can denote \( R_1 = R \) and \( R_2 = -R \). Thus, we have:

\[ \frac{1}{f} = \left( \frac{\frac{4}{3} - 1}{R} + \frac{\frac{4}{3} - 1}{R} \right) = \frac{2 \left(\frac{4}{3} - 1\right)}{R} \]

Calculating \( \frac{4}{3} - 1 = \frac{1}{3} \):

\[ \frac{1}{f} = \frac{2 \cdot \frac{1}{3}}{R} = \frac{2}{3R} \]

Thus, we can express this as:

\[ \frac{2}{R} = \frac{3}{f} \quad \text{(Equation 1)} \]

### Step 3: Calculate the focal length of the liquid lens in glass
Now, we need to find the focal length when the lens is placed in glass. Given:
- Refractive index of glass, \( \mu_g = \frac{3}{2} \)

Using the lens maker's formula again for the lens in glass:

\[ \frac{1}{f'} = \left( \frac{\frac{4}{3} - \frac{3}{2}}{R_1} - \frac{\frac{4}{3} - \frac{3}{2}}{R_2} \right) \]

Calculating \( \frac{4}{3} - \frac{3}{2} \):

Finding a common denominator (6):

\[ \frac{4}{3} = \frac{8}{6}, \quad \frac{3}{2} = \frac{9}{6} \quad \Rightarrow \quad \frac{4}{3} - \frac{3}{2} = \frac{8}{6} - \frac{9}{6} = -\frac{1}{6} \]

So, we have:

\[ \frac{1}{f'} = \left( \frac{-\frac{1}{6}}{R} + \frac{\frac{1}{6}}{R} \right) = \frac{-2 \cdot \frac{1}{6}}{R} = -\frac{1}{3R} \]

### Step 4: Relate \( f' \) to \( f \)
From Equation 1, we know \( \frac{2}{R} = \frac{3}{f} \). Therefore, \( R = \frac{3f}{2} \).

Substituting this into the equation for \( f' \):

\[ \frac{1}{f'} = -\frac{1}{3 \cdot \frac{3f}{2}} = -\frac{2}{9f} \]

Thus, we have:

\[ f' = -\frac{9f}{2} \]

### Step 5: Determine the nature of the lens
Since \( f' \) is negative, this indicates that the lens behaves like a concave lens when placed in glass.

### Final Answer
The focal length of the liquid lens in glass is \( -\frac{9f}{2} \) and the nature of the lens is concave.

---