Question

A student can distinctly see the object up to a distance $15\text{cm}.$ He wants to see the black board at a distance of $3\text{m}.$ Focal length and power of lens used respectively will be-

• A. $–4.8\text{cm};–3.3\text{D}$
• B. $–5.8\text{cm};–4.3\text{D}$
• C. $–7.5\text{cm};–6.3\text{D}$
• D. $–15.8\text{cm};–6.3\text{D}$

Answer 1

The correct option is D $–15.8\text{cm};–6.3\text{D}$

As student can see only upto $15\text{cm}$ so if he wants to see black board which is at distance of $3\text{m}$ he has to use a lens which can make an image of black board at $15\text{cm}.$

$v=-15\text{cm};u=-300\text{cm}$

Using lens formula,

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

$\Rightarrow \frac{1}{f}=\frac{1}{-15}-\frac{1}{-300}$

$\Rightarrow f=\frac{-300}{19}\approx -15.8\text{cm}$

Power of the lens $P=\frac{1}{f}$

Where, $f=\text{focal length in meter}$

$\Rightarrow P=-\frac{19\times 100}{300}=\frac{-19}{3}=-6.33\text{D}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.