Question

a person cannot see the object distinctly when placed at a distance less than 50cm
a) identify the defect of vision
b) calculate the power and nature of the lens he should be using to see clearly the object placed at a distance of 25cm from his eyes

Answer 1

Hypermetropia is defect of vision in which the person cannot see object when placed at distance less than 50 cm.

Power of lens is 2D.

Solution:

Lens maker formula, $-\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$  

Given:

Distance from object u = - 25 cm  

Distance from image = - 50 cm

Applying formula, $\frac{1}{f}=\frac{1}{(-50)}-\frac{1}{(-25)}$

$\begin{array}{l}{\frac{1}{f}=-\frac{1}{50}+\frac{1}{25}} \\ \\{\frac{1}{f}=\frac{-1+2}{50}} \\ \\{\frac{1}{f}=\frac{1}{50}}\end{array}$

f= +50 cm

Now that the given focal length has come up to be in positive figures, implies that lens must be used be of convex lens.

Power of convex =$\frac{1}{f}$

Power= $\frac{1}{0.5\ m}$

Power = 2 D

Answer 2

Answer:

Explanation:

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Answered

A person cannot see the objects distinctly when placeda at a distance less than 50 cm. Calculate the power and the nature of the lens he should be using to see clearly the object placed at a distance of 25 cm from his eyes..

Asked by Anukruti Valase | 27th Feb, 2015,  04:32: PM

Expert Answer:

Given that the person can see distant objects clearly but not near by objects. This defect of eye is known as long sightedness or hypermetropia.

Given that:

v=- 50 cm , least distance of distinct vision, u =-25 cm

According to lens formula: