• # question_answer 34) A student can distinctly see the object upto a distance 15 cm. He wants to see the black board at a distance  of 3 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)  $-15.8\text{ }cm,-6.3\text{ }D$

$v=-15\,cm,$ $u=-300\,cm$ from lens formula $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ $\Rightarrow$ $\frac{1}{f}=\frac{1}{-15}-\frac{1}{-300}$ $\Rightarrow$ $f=\frac{-300}{19}=-15.8\,cm$ and power $P=\frac{100}{f}cm=\frac{-100\times 19}{300}=-6.33D$