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Q.
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 :
JIPMERJIPMER 2004Ray Optics and Optical Instruments
Solution:
The student should use a lens which forms image at distance of $15\, cm$ of the object placed at $3\, m$ i.e., object distance $u=-3\, m =-300\, cm$, image distance $v=-15\, cm$
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 $f=\frac{-300}{19}=-15.8\, cm$ (concave lens)
Now power of the lens is
$P=\frac{100}{f( cm )}=\frac{100}{-015.8}=-6.33\, D$