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

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) (10/3) cm (B) 30 cm (C) 15 cm (D) (100/3) cm

Tardigrade Admin · Accepted Answer

For improving far point, concave lens is required and for this concave lens u=∞, v=-30 cm So (1/f)=(1/-30)-(1/∞) ⇒ f=-30 cm for near point (1/-30)=(1/-15)-(1/-15)-(1/u) ⇒ u=-30 cm

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

$\frac{10}{3} c m$

$30 c m$

$15 c m$

$\frac{100}{3} c m$

For improving far point, concave lens is required and for this concave lens
$u = \infty, v = - 30 c m$
So $\frac{1}{f} = \frac{1}{- 30} - \frac{1}{\infty}$
$\Rightarrow f = - 30 c m$
for near point $\frac{1}{- 30} = \frac{1}{- 15} - \frac{1}{- 15} - \frac{1}{u}$
$\Rightarrow u = - 30 c m$

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

310​cm

30cm

15cm

3100​cm