Question

a person wants to read a book placed at 15 CM whereas near point of his eye is 30cm the power of required lens is​

Answer 1

Answer:

The answer will be 0.66D and the defect is Hypermetropia

Explanation:

we khow that,

XD/X-D

wher X=30cm and D=25cm

therefore.. (f) =150cm

P=1/f

p=1/150 and thw focal lenght should always be in meters sooo....

p=1×100/150

=0.66D

Answer 2

The power of required lens is 3.33 D.

Explanation:

Given that,

Object distance u= -30 cm

Image distance v= -15 cm

We need to calculate the focal length

Using formula of lens

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

Where, u = object distance

v = image distance

Put the value into the formula

$\dfrac{1}{f}=\dfrac{1}{-15}-\dfrac{1}{-30}$

$\dfrac{1}{f}=\dfrac{1}{-15}+\dfrac{1}{30}$

$f=-30\ cm$

We need to calculate the power

Using formula of power

$P=\dfrac{100}{f}$

$P=\dfrac{100}{30}$

$P=3.33\ D$

Hence, The power of required lens is 3.33 D.

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Topic : Power of lens

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