Question

A professor reads a greeting card received on his ${50}^{th}$ birthday with $+2.5\text{D}$ glass keeping the card $25\text{cm}$ away. Ten years later, when he reads his farewell letter with the same glasses, he has to keep the letter $50\text{cm}$ away. The power of lens the professor should use now is

• A. $3\text{D}$
• B. $4\text{D}$
• C. $4.5\text{D}$
• D. $6\text{D}$

Answer 1

The correct option is C $4.5\text{D}$

The professor reads the card at a distance $25\text{cm}$ using a glass of $+2.5\text{D}$.

Since power of a lens is reciprocal of its focal length, we can write that
$f=\frac{1}{2.5}=40\text{cm}$

After ten years, when he reads farewell letter, he keeps the letter at a distance $50\text{cm}$.
$u=-50\text{cm}$

Now using lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}=\frac{1}{u}+\frac{1}{f}$

Substituting the given data we get,

$\frac{1}{v}=-\frac{1}{50}+\frac{1}{40}$

$\Rightarrow v=200\text{cm}$

To read the farewell letter at a distance of $25\text{cm}$ [near point]

$u=-25\text{cm}$

Again using lens formula,

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{f}=\frac{1}{200}-\frac{-1}{-25}\Rightarrow \frac{1}{f}=\frac{9}{200}$

$\Rightarrow f=\frac{2}{9}\text{m}$

Thus, required power of lens $=\frac{1}{f}=\frac{9}{2}=4.5\text{D}$.

Hence, option (c) is the correct answer.