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Question: A student can distinctly see the object upto a distance 15 cm. He wants to see the black board at a ...

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

−4.8cm, 6mu−3.3D- 4.8cm,\mspace{6mu} - 3.3D

B

−5.8cm, 6mu−4.3D- 5.8cm,\mspace{6mu} - 4.3D

C

−7.5cm, 6mu−6.3D- 7.5cm,\mspace{6mu} - 6.3D

D

−15.8cm, 6mu−6.3D- 15.8cm,\mspace{6mu} - 6.3D

Answer

−15.8cm, 6mu−6.3D- 15.8cm,\mspace{6mu} - 6.3D

Explanation

Solution

v=−15cm,v = - 15cm, u=−300cm,u = - 300cm,

From lens formula 1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

⇒1f=1−15−1−300=−19300⇒f=−30019=−15.8cm\Rightarrow \frac{1}{f} = \frac{1}{- 15} - \frac{1}{- 300} = \frac{- 19}{300} \Rightarrow f = \frac{- 300}{19} = - 15.8cm

and power P=100fcm=−100×19300P = \frac{100}{f}cm = \frac{- 100 \times 19}{300}= ­– 6.33 D.