First slide

The optical instruments

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

Moderate

A

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

30 cm
C

15 cm
D

$\frac{100}{3} cm$

Solution

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

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