A professor reads a greeting card received on his 50th birthday with +2.5D glasses keeping the card 25cm away. Ten years later, he reads his farewell letter with the same glasses but he has to keep the letter 50cm away. What power of lens should he now use?
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On the 50th birthday, he reads the card at a distance 25cm using a glass of +2.5D.Ten years later, his near point must have changed.
So after ten years,
$\text{u}=-50\text{cm},$
$\text{f}=\frac{1}{2.5\text{D}}=0.4\text{m}=40\text{cm}$
$\text{v}=\text{near point}$
Now, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow \frac{1}{\text{v}}=\frac{1}{\text{u}}+\frac{1}{\text{f}}=\frac{1}{-50}+\frac{1}{40}=\frac{1}{200}$
So, near point = v = 200cm
To read the farewell letter at a distance of 25cm,
U = –25cm
For lens formula,
$\Rightarrow\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{200} -\frac{1}{-25}=\frac{1}{200}+\frac{1}{25}=\frac{9}{200}$
$\Rightarrow\text{f}=\frac{200}{9}\text{cm}=\frac{2}{9}\text{m}$
⇒ Power of the lens $=\frac{1}{\text{f}}=\frac{9}{2}=4.5\text{D.}$
$\therefore$ He has to use a lens of power +4.5D.
So after ten years,
$\text{u}=-50\text{cm},$
$\text{f}=\frac{1}{2.5\text{D}}=0.4\text{m}=40\text{cm}$
$\text{v}=\text{near point}$
Now, $\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow \frac{1}{\text{v}}=\frac{1}{\text{u}}+\frac{1}{\text{f}}=\frac{1}{-50}+\frac{1}{40}=\frac{1}{200}$
So, near point = v = 200cm
To read the farewell letter at a distance of 25cm,
U = –25cm
For lens formula,
$\Rightarrow\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
$\Rightarrow\frac{1}{\text{f}}=\frac{1}{200} -\frac{1}{-25}=\frac{1}{200}+\frac{1}{25}=\frac{9}{200}$
$\Rightarrow\text{f}=\frac{200}{9}\text{cm}=\frac{2}{9}\text{m}$
⇒ Power of the lens $=\frac{1}{\text{f}}=\frac{9}{2}=4.5\text{D.}$
$\therefore$ He has to use a lens of power +4.5D.
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