Question

A man can see the object between 15 cm and 30 cm. He uses the lens to see the far objects. Then due to the lens used, the near point will be at

• A. $\frac{10}{3}$cm
• B. 30 cm
• C. 15 cm
• D. $\frac{100}{3}$cm

Answer 1

Answer: (B)

30 cm

Answer 2

For improving far point, concave lens is required and for this concave lens $u = \infty,\mspace{6mu}\mspace{6mu} v = - \mspace{6mu} 30\mspace{6mu} cm$

So $\frac{1}{f} = \frac{1}{- 30} - \frac{1}{\infty} \Rightarrow f = - \mspace{6mu} 30\mspace{6mu} cm$

for near point $\frac{1}{- 30} = \frac{1}{- 15} - \frac{1}{u} \Rightarrow u = - \mspace{6mu} 30\mspace{6mu} cm$