Question

A student can distinctly see the object upto a distance $15cm$. He wants to see the black board at a distance of $3m$. Focal length and power of lens used respectively will be :

• A. -4.8 cm ,-3.3 D
• B. -5.8 cm ,-4.3 D
• C. -7.5 cm ,-6.3 D
• D. -15.8 cm ,-6.33 D

Answer 1

Answer: (D)

The student should use a lens which forms image at distance of $15cm$ of the object placed at $3m$ i.e., object distance $u=-3m=-300cm$, image distance $v=-15cm$
From lens formula $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
we get $\frac{1}{-15}-\frac{1}{-300}=\frac{1}{f}$
or $\frac{1}{f}=\frac{1}{300}-\frac{1}{15}=-\frac{19}{300}$
or $f=\frac{-300}{19}=-15.8cm$ (concave lens)
Now power of the lens is
$P=\frac{100}{f(cm)}=\frac{100}{-015.8}=-6.33D$