A thin biconvex lens of focal length 6.25 cm is made of material of refractive index 1.5. It is cut into two identical pieces perpendicular to its principal axis. One of the pieces is placed in water of refractive index `4/3`. The focal power of the piece immersed in water in diopter is

Answer

To find the focal power of the piece of the biconvex lens immersed in water, we will follow these steps:

### Step 1: Understand the problem
We have a biconvex lens with a focal length of 6.25 cm and a refractive index of 1.5. When cut into two pieces, one piece is placed in water (refractive index = 4/3). We need to find the focal power of the piece in water.

### Step 2: Use the Lens Maker's Formula
The Lens Maker's Formula is given by:
\[ \frac{1}{f} = \left( n_2 - n_1 \right) \cdot \frac{1}{n_1} \left( \frac{1}{r_1} - \frac{1}{r_2} \right) \]
Where:
- \( f \) = focal length of the lens
- \( n_1 \) = refractive index of the medium (air = 1)
- \( n_2 \) = refractive index of the lens material (1.5)
- \( r_1 \) and \( r_2 \) are the radii of curvature of the lens.

### Step 3: Determine the radii of curvature
Since the lens is biconvex and symmetric, we can assume:
- \( r_1 = r \) (the radius of the convex surface)
- \( r_2 = -r \) (the radius of the other convex surface, taken as negative)

Using the given focal length:
\[ \frac{1}{6.25} = (1.5 - 1) \cdot \frac{1}{1} \left( \frac{1}{r} - \frac{1}{-r} \right) \]
\[ \frac{1}{6.25} = 0.5 \cdot \left( \frac{2}{r} \right) \]
\[ \frac{1}{6.25} = \frac{1}{r} \]
Thus, \( r = 6.25 \) cm.

### Step 4: Calculate the new focal length in water
Now, when the lens is cut, one piece has one flat surface (infinite radius of curvature) and the other surface with radius \( r = 6.25 \) cm. The new focal length \( f' \) when the lens is in water can be calculated using the modified Lens Maker's Formula:
\[ \frac{1}{f'} = \left( n_2 - n_1 \right) \cdot \frac{1}{n_1} \left( \frac{1}{r_1} - \frac{1}{\infty} \right) \]
Where:
- \( n_1 = \frac{4}{3} \) (refractive index of water)
- \( n_2 = 1.5 \) (refractive index of the lens)
- \( r_1 = 6.25 \) cm

Substituting the values:
\[ \frac{1}{f'} = \left( 1.5 - \frac{4}{3} \right) \cdot \frac{1}{\frac{4}{3}} \cdot \frac{1}{6.25} \]
Calculating \( 1.5 - \frac{4}{3} = \frac{1.5 \cdot 3 - 4}{3} = \frac{4.5 - 4}{3} = \frac{0.5}{3} \):
\[ \frac{1}{f'} = \left( \frac{0.5}{3} \right) \cdot \left( \frac{3}{4} \right) \cdot \frac{1}{6.25} \]
\[ \frac{1}{f'} = \frac{0.5}{4 \cdot 6.25} = \frac{0.5}{25} = \frac{1}{50} \]
Thus, \( f' = 50 \) cm.

### Step 5: Calculate the focal power in diopters
Focal power \( P \) in diopters is given by:
\[ P = \frac{1}{f'} \text{ (in meters)} \]
Converting \( f' \) from cm to meters:
\[ f' = 0.50 \text{ m} \]
Thus,
\[ P = \frac{1}{0.50} = 2 \text{ diopters} \]

### Final Answer
The focal power of the piece immersed in water is **2 diopters**.