A thin convex lens of refractive index 1.5cm has 20cm focal length in air. If the lens in completely immersed in a liquid of refractive index. 1.6, then its focal length will be

Options

A. `-160cm`
B. `-100cm`
C. `+10cm`
D. `+100cm`

Explanation

To solve the problem of finding the new focal length of a thin convex lens when it is immersed in a liquid, we can use the lensmaker's formula. Here’s a step-by-step solution:

### Step 1: Understand the Lensmaker's Formula
The lensmaker's formula relates the focal length of a lens to its radii of curvature and the refractive indices of the lens material and the surrounding medium. The formula is given by:
\[ \frac{1}{F} = \left( n - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
where:
- \( F \) is the focal length of the lens,
- \( n \) is the refractive index of the lens material,
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

### Step 2: Identify Given Values
From the problem:
- Refractive index of the lens, \( n = 1.5 \)
- Focal length in air, \( F_1 = 20 \, \text{cm} \)
- Refractive index of the liquid, \( n_{liquid} = 1.6 \)

### Step 3: Calculate the Radii of Curvature
Using the lensmaker's formula in air:
\[ \frac{1}{F_1} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the known values:
\[ \frac{1}{20} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
\[ \frac{1}{20} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
This implies:
\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{10} \]

### Step 4: Apply the Lensmaker's Formula in Liquid
Now, when the lens is immersed in the liquid, we need to use the modified lensmaker's formula:
\[ \frac{1}{F_2} = \left( n_{glass} - n_{liquid} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Substituting the values:
\[ \frac{1}{F_2} = (1.5 - 1.6) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
\[ \frac{1}{F_2} = (-0.1) \left( \frac{1}{10} \right) \]
\[ \frac{1}{F_2} = -0.01 \]

### Step 5: Calculate the New Focal Length
Taking the reciprocal gives:
\[ F_2 = \frac{1}{-0.01} = -100 \, \text{cm} \]

### Conclusion
The new focal length of the lens when immersed in the liquid is \( F_2 = -100 \, \text{cm} \).