Question

A person with a defective eye vision is unable to see the objects nearer than 1.5m. He wants to read books at a distance of 30cm. Find the nature, focal length and power of the lens he needs in his spectacles

Answer 1

Answer:0.7cm

Explanation:

Answer 2

Hypermetropia.

The nature, focal length and power of the lens are convex, 37.5 cm and 2.67 D respectively.

Explanation:

The person sufferes from defect of hypermetropia( long-sightedness).

The object distance, u  = -30 cm.

The image distance , v = -1.5 m = -150 cm.

Use lens formula as,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Here, f is the focal length.

substitute the values, we get

$\frac{1}{f}=\frac{1}{-150cm}-\frac{1}{-30cm}$

$\frac{1}{f}=\frac{150cm}{4}$

f = 37.5 cm.

Therefore, power of the lens,

$P=\frac{1}{f(m)}$

so,

$P=\frac{100}{37.5}$

P = 2.67 D.

As focal length is positive so, the person needed convex lens.

Thus the nature, focal length and power of the lens are convex, 37.5 cm and 2.67 D respectively.

#Learn More: lens formula.

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