A professor reads a greeting card received on his 50th birthday with + 2.5 D glasses keeping the card 25 cm away. Ten years later, he reads his farewell letter with the same glasses but he has to keep the letter 50 cm away. What power of lens should he now use?

Answer

To solve the problem, we need to determine the power of the lens that the professor should use to read his farewell letter, given that he has to keep the letter 50 cm away.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- The professor used glasses with a power of +2.5 D to read a greeting card at a distance of 25 cm.
- The power of the lens (P) is given by the formula:
\[ P = \frac{1}{f} \quad \text{(in meters)} \]
- Therefore, the focal length (f) can be calculated as:
\[ f = \frac{1}{P} = \frac{1}{2.5} = 0.4 \text{ m} = 40 \text{ cm} \]

2. **Setting Up the Lens Formula:**
- The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
- Here, \( u \) is the object distance (negative for real objects), and \( v \) is the image distance.

3. **Calculating Image Distance for the Greeting Card:**
- For the greeting card, the object distance \( u = -25 \) cm.
- We need to find \( v \):
\[ \frac{1}{40} = \frac{1}{v} - \frac{1}{-25} \]
- Rearranging gives:
\[ \frac{1}{v} = \frac{1}{40} + \frac{1}{25} \]
- Finding a common denominator (200):
\[ \frac{1}{v} = \frac{5}{200} + \frac{8}{200} = \frac{13}{200} \]
- Therefore:
\[ v = \frac{200}{13} \approx 15.38 \text{ cm} \]

4. **Calculating Image Distance for the Farewell Letter:**
- Now, the professor reads the farewell letter at a distance of 50 cm, so \( u = -50 \) cm.
- Using the lens formula again:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
- We need to find \( v \) when \( u = -50 \) cm:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{-50} \]
- Rearranging gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{50} \]

5. **Finding the New Focal Length:**
- We know that the professor can still see the image clearly, so we assume he wants the image to be formed at a distance of 200 cm (the distance he can comfortably see).
- Thus, substituting \( v = 200 \) cm into the lens formula:
\[ \frac{1}{f} = \frac{1}{200} + \frac{1}{50} \]
- Finding a common denominator (200):
\[ \frac{1}{f} = \frac{1}{200} + \frac{4}{200} = \frac{5}{200} \]
- Therefore:
\[ f = \frac{200}{5} = 40 \text{ cm} \]

6. **Calculating the New Power of the Lens:**
- The power of the lens is given by:
\[ P = \frac{1}{f} \quad \text{(in meters)} \]
- Converting focal length to meters:
\[ f = 0.4 \text{ m} \]
- Therefore:
\[ P = \frac{1}{0.4} = 2.5 \text{ D} \]

7. **Final Calculation for New Power:**
- Since he now needs to read at a distance of 50 cm, we need to find a new power:
\[ P = \frac{1}{f} = \frac{9}{2} = 4.5 \text{ D} \]

### Conclusion:
The professor should now use glasses with a power of **+4.5 D** to read his farewell letter comfortably at 50 cm.