A student can distinctly see the object upto a distance 15 cm. He...

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 :

Options

A $-4.8 \mathrm{~cm},-3.3 \mathrm{D}$
B $-5.8 \mathrm{~cm},-4.3 \mathrm{D}$
C $-7.5 \mathrm{~cm},-6.3 \mathrm{D}$
D $-15.8 \mathrm{~cm},-6.33 \mathrm{D}$

Correct answer

D. $-15.8 \mathrm{~cm},-6.33 \mathrm{D}$

Step-by-step 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 \mathrm{~m}=-300 \mathrm{~cm}

, image distance

v=-15 \mathrm{~cm}

From lens formula

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

we get

\quad \frac{1}{-15}-\frac{1}{-300}=\frac{1}{f}

\quad \frac{1}{f}=\frac{1}{300}-\frac{1}{15}=-\frac{19}{300}

\quad f=\frac{-300}{19}=-15.8 \mathrm{~cm} \quad

(concave lens) Now power of the lens is

P=\frac{100}{f(\mathrm{~cm})}=\frac{100}{-15.8}=-6.33 \mathrm{D}

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