A concave lens with unequal radii of curvature made of glass `(mu_(g)=1.5)` has a focal length of 40cm. If it is immersed in a liquid of refractive index `mu_(1)=2`, then

Options

A. it behaves like a convex lens of 80cm focal length
B. it behaves like a convex lens of 20cm focal length
C. its focal length becomes 60cm
D. nothing can be said

Explanation

To solve the problem step by step, we will analyze the given information and apply the lens maker's formula to find the new focal length of the concave lens when immersed in a liquid.

### Step 1: Understand the given data
- The refractive index of glass, \( \mu_g = 1.5 \)
- The focal length of the concave lens in air, \( f = -40 \, \text{cm} \) (negative because it is a concave lens)
- The refractive index of the liquid, \( \mu_1 = 2 \)

### Step 2: Use the lens maker's formula in air
The lens maker's formula for a lens in air is given by:
\[ \frac{1}{f} = \left( \mu_g - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
For the concave lens, we know:
\[ \frac{1}{-40} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
This simplifies to:
\[ \frac{1}{-40} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Thus, we can express:
\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{-20} \]

### Step 3: Calculate the new focal length in the liquid
When the lens is immersed in a liquid, we use the modified lens maker's formula:
\[ \frac{1}{f'} = \left( \frac{\mu_g}{\mu_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{f'} = \left( \frac{1.5}{2} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Calculating \( \frac{1.5}{2} \):
\[ \frac{1.5}{2} = 0.75 \]
So we have:
\[ \frac{1}{f'} = (0.75 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
This simplifies to:
\[ \frac{1}{f'} = -0.25 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting \( \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{-20} \):
\[ \frac{1}{f'} = -0.25 \left( \frac{1}{-20} \right) \]
Calculating this gives:
\[ \frac{1}{f'} = \frac{0.25}{20} = \frac{1}{80} \]

### Step 4: Find the new focal length
Thus, we find:
\[ f' = 80 \, \text{cm} \]
Since the focal length is positive, the lens behaves like a convex lens in the liquid.

### Final Answer
The new focal length of the concave lens when immersed in the liquid is \( f' = 80 \, \text{cm} \).

---