A professor reads a greeting card received on his 50 th birthday...

Using the lens formula in terms of power, $P = \dfrac{1}{v} - \dfrac{1}{u}$.

At age $60$, the professor uses the same glasses with power $P = +2.5 \text{ D}$ to read the letter at a distance $u = -50 \text{ cm} = -0.5 \text{ m}$. The image is formed at his new near point $v$.

Substituting the values to find the vergence of his near point:

$\dfrac{1}{v} = P + \dfrac{1}{u}$

$\dfrac{1}{v} = 2.5 + \left(\dfrac{1}{-0.5}\right)<br><br>\dfrac{1}{v} =$2.5 - 2.0 = 0.5 $\text{ D}<br><br>To read at the normal near point distance of$25 $\text{ cm} ($u' = -25 $\text{ cm} = -$0.25 $\text{ m}), the new lens must form an image at this same near point$v$.<br><br>Let the new power required be$P'$. Using the lens formula again:<br><br>$P' = $\dfrac{1}{v} - \dfrac{1}{u'}<br><br>$P' = 0.5 - \left(\dfrac{1}{-0.25}\right)$<br><br>$P' = 0.5 - (-4.0) = +4.5 \text{ D}$<br><br>Therefore, the power of the lens he should now use is$+4.5 \text{ D}$.<br><br>Answer:$+4.5 \text{ D}