Question

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 1

Given:
The professor can read the card at a distance of 25 cm using a glass of + 2.5 D.
After 10 years, he can read the farewell letter at a distance (u) of – 50 cm.
Focal length, f = $\frac{1}{\mathrm{Power}}$ = $\frac{1}{2.5}$ = 0.4 m = 40 cm
The lens formula is given by
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Putting the values, we get:
$$\begin{gathered} \frac{1}{v}=\frac{1}{\left(-50\right)}+\frac{1}{40} \\ \Rightarrow v=200\mathrm{cm} \end{gathered}$$
∴ His near point, v = 200 cm
To read the farewell letter at a distance of 25 cm,
u = – 25 cm
The lens formula is given by
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
Putting the values, we get:
$$\begin{gathered} \frac{1}{f}=\frac{1}{\left(200\right)}-\frac{1}{\left(-25\right)}=+\frac{9}{200} \\ \Rightarrow f=\frac{200}{9}\mathrm{cm}=\frac{2}{9}\mathrm{m} \end{gathered}$$
∴ Power of the lens = $\frac{1}{f}$ = 4.5 D
He should use a lens of power + 4.5 D.