The near point of a person is 75 cm. In order that he may be able to read book at a distance 30 cm. The power of spectacles lenses should be

Options

A. `-2D`
B. `+3.75D`
C. `+2D`
D. `+3D`

Answer

To determine the power of the spectacles lenses needed for a person whose near point is 75 cm to read a book at a distance of 30 cm, we can follow these steps:

### Step 1: Understand the Problem
The near point of the person is 75 cm, meaning he can see objects clearly only at distances greater than or equal to 75 cm. To read a book at 30 cm, we need to use a lens that will allow him to see this distance as if it were at his near point.

### Step 2: Define the Variables
- The object distance (u) is the distance of the book from the lens, which is -30 cm (negative because it is on the same side as the object).
- The image distance (v) is the distance at which the person can see clearly, which is 75 cm (positive because it is on the opposite side of the lens).

### Step 3: Use the Lens Formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Where:
- \( f \) is the focal length of the lens,
- \( v \) is the image distance,
- \( u \) is the object distance.

### Step 4: Substitute the Values
Substituting the values into the lens formula:
\[ \frac{1}{f} = \frac{1}{75} - \frac{1}{-30} \]
This simplifies to:
\[ \frac{1}{f} = \frac{1}{75} + \frac{1}{30} \]

### Step 5: Find a Common Denominator
To add the fractions, find a common denominator. The least common multiple of 75 and 30 is 150. Thus:
\[ \frac{1}{75} = \frac{2}{150}, \quad \frac{1}{30} = \frac{5}{150} \]
Now, substituting these values:
\[ \frac{1}{f} = \frac{2}{150} + \frac{5}{150} = \frac{7}{150} \]

### Step 6: Calculate the Focal Length
Taking the reciprocal gives:
\[ f = \frac{150}{7} \approx 21.43 \text{ cm} \]

### Step 7: Calculate the Power of the Lens
The power \( P \) of a lens is given by:
\[ P = \frac{1}{f} \text{ (in meters)} \]
First, convert the focal length from cm to meters:
\[ f = \frac{150}{7} \text{ cm} = \frac{150}{700} \text{ m} = \frac{15}{70} \text{ m} \approx 0.2143 \text{ m} \]
Now, calculate the power:
\[ P = \frac{1}{0.2143} \approx 4.67 \text{ diopters} \]

### Step 8: Conclusion
Since the power is positive, it indicates that the lens is a convex lens. Therefore, the power of the spectacles lenses required is approximately 4.67 diopters.