A person has near point at 100 cm. What power of lens is needed to read at 20 cm if he/she uses (a) contact lens, (b) spectacles having glasses 2.0 cm separated from the eyes?

Answer

To solve the problem, we need to determine the power of the lens required for a person with a near point at 100 cm to read at 20 cm. We will consider two cases: (a) using contact lenses and (b) using spectacles with a distance of 2 cm from the eyes.

### Step-by-Step Solution

#### Part (a): Using Contact Lens

1. **Identify the Object and Image Distances**:
- The object distance (u) is the distance from the lens to the object, which is 20 cm. According to the sign convention, we take this as negative:
\[ u = -20 \text{ cm} = -0.2 \text{ m} \]
- The image distance (v) is the distance from the lens to the image, which is the near point of the person (100 cm). This is also taken as negative:
\[ v = -100 \text{ cm} = -1 \text{ m} \]

2. **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 \(u\):
\[ \frac{1}{f} = \frac{1}{-1} - \frac{1}{-0.2} \]
\[ \frac{1}{f} = -1 + 5 = 4 \]

3. **Calculate the Focal Length**:
\[ f = \frac{1}{4} \text{ m} = 0.25 \text{ m} \]

4. **Calculate the Power of the Lens**:
The power \(P\) of a lens is given by:
\[ P = \frac{1}{f} \text{ (in meters)} \]
\[ P = \frac{1}{0.25} = 4 \text{ diopters (D)} \]

#### Part (b): Using Spectacles

1. **Identify the Object and Image Distances**:
- The object distance (u) is now adjusted because the spectacles are 2 cm away from the eyes. Therefore:
\[ u = - (20 \text{ cm} - 2 \text{ cm}) = -18 \text{ cm} = -0.18 \text{ m} \]
- The image distance (v) remains the same:
\[ v = -100 \text{ cm} = -1 \text{ m} \]

2. **Use the Lens Formula**:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values of \(v\) and \(u\):
\[ \frac{1}{f} = \frac{1}{-1} - \frac{1}{-0.18} \]
\[ \frac{1}{f} = -1 + \frac{100}{18} = -1 + 5.56 \approx 4.56 \]

3. **Calculate the Focal Length**:
\[ f \approx \frac{1}{4.56} \text{ m} \approx 0.219 \text{ m} \]

4. **Calculate the Power of the Lens**:
\[ P = \frac{1}{f} \approx 4.56 \text{ diopters (D)} \]

### Final Answers
- (a) Power of contact lens: **4 D**
- (b) Power of spectacles: **4.56 D**