The near point and far point of a person are 40cm and 250cm, respectively. Determine the power of lens he/she should use while reading a bool kept at distance 25cm away from it,

Options

A. 2.5D
B. 5D
C. 1.5D
D. 3.5D

Explanation

To determine the power of the lens required for a person whose near point is 40 cm and far point is 250 cm, we will follow these steps:

### Step 1: Identify the given values
- Near point (N) = 40 cm = 0.4 m
- Far point (F) = 250 cm = 2.5 m
- Distance of the book (d) = 25 cm = 0.25 m

### Step 2: Determine the image distance (v)
The lens should be placed such that the image of the object (the book) appears at the near point of the person. Therefore, the image distance (v) will be equal to the near point:
- v = -N = -0.4 m (the negative sign indicates that the image is virtual and on the same side as the object).

### Step 3: Determine the object distance (u)
The object distance (u) is the distance of the book from the lens:
- u = -d = -0.25 m (the negative sign indicates that the object is on the same side as the incoming light).

### Step 4: Use the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values of v and u:
\[ \frac{1}{f} = \frac{1}{-0.4} - \frac{1}{-0.25} \]

### Step 5: Calculate the values
Calculating the right-hand side:
\[ \frac{1}{f} = -\frac{1}{0.4} + \frac{1}{0.25} \]
Finding a common denominator (which is 1):
\[ \frac{1}{f} = -\frac{2.5}{1} + \frac{4}{1} = -2.5 + 4 = 1.5 \]
Thus, we have:
\[ f = \frac{1}{1.5} \approx 0.6667 \text{ m} \]

### Step 6: Calculate the power of the lens
The power (P) of the lens is given by:
\[ P = \frac{1}{f} \text{ (in meters)} \]
Substituting the value of f:
\[ P = \frac{1}{0.6667} \approx 1.5 \text{ D} \]

### Final Answer
The power of the lens required is approximately **1.5 Diopters**.

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