A bi-convex lens made of glass (refractive index 1.5) is put in a liquid of refractive index 1.7. Its focal length will

Options

A. Decrease and change sign
B. Increase and change sign
C. Decrease and remain of the same sign
D. Increase and remain of the same sign

Answer

To find the focal length of a bi-convex lens made of glass (refractive index 1.5) when placed in a liquid with a refractive index of 1.7, we can use the lensmaker's formula. Let's go through the steps systematically.

### Step 1: Understand the Lensmaker's Formula
The lensmaker's formula is given by:

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

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

### Step 2: Calculate the Focal Length in Air
When the lens is in air (refractive index = 1), the relative refractive index \( \mu \) is:

\[ \mu_{air} = \frac{n_{lens}}{n_{air}} = \frac{1.5}{1} = 1.5 \]

Substituting into the lensmaker's formula:

\[ \frac{1}{f_{air}} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

This simplifies to:

\[ \frac{1}{f_{air}} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Let’s denote \( k = \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \), then:

\[ \frac{1}{f_{air}} = 0.5k \quad \Rightarrow \quad f_{air} = \frac{2}{k} \]

### Step 3: Calculate the Focal Length in the Liquid
Now, when the lens is placed in a liquid with a refractive index of 1.7, the relative refractive index \( \mu \) becomes:

\[ \mu_{liquid} = \frac{n_{lens}}{n_{liquid}} = \frac{1.5}{1.7} \]

Substituting into the lensmaker's formula:

\[ \frac{1}{f_{liquid}} = \left( \frac{1.5}{1.7} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Calculating \( \frac{1.5}{1.7} - 1 \):

\[ \frac{1.5}{1.7} - 1 = \frac{1.5 - 1.7}{1.7} = \frac{-0.2}{1.7} \]

Thus:

\[ \frac{1}{f_{liquid}} = \frac{-0.2}{1.7} k \]

This gives:

\[ f_{liquid} = \frac{1.7}{-0.2} \cdot \frac{1}{k} = -8.5 \cdot \frac{1}{k} \]

### Step 4: Compare Focal Lengths
From our calculations:
- \( f_{air} = \frac{2}{k} \)
- \( f_{liquid} = -8.5 \cdot \frac{1}{k} \)

We can see that the focal length has changed from a positive value in air to a negative value in the liquid, indicating that the lens now behaves as a diverging lens in the liquid.

### Conclusion
The focal length of the bi-convex lens when placed in a liquid of refractive index 1.7 will be negative, indicating that it has become a diverging lens.