Question

how can you decide the focal length of the lens to be used to convert hypermetropia ?deduce mathematically? ​

Answer 1

Answer:

Calculation of focal length and power of correcting lens in hypermetropia: Let y = distance of the near point N' from the defective eye. Now, the near point N of the normal eye is at distance D = 25 cm. The object placed at N forms its virtual image at N' due to the convex lens used for correction of the defect.

Answer 2

Answer:

Given: The focal length of a lens suggested to a person with Hypermetropia is 100cm.

To find the distance of near point and power of the lens.

Solution:

Let the distance of near point be ’d’ and focal length be 'f=100cm' then

f=

d−25

25d

cm

⟹100(d−25)=25d

⟹100d−25d=2500

⟹d=

75

2500

⟹d=

3

100

=33.33cm

Hence the distance of near point will be 33.33cm

We know,

Power, P=

f

1

D, f is in metres

⟹P=

100×10

−2

1

⟹P=1D

is the power of the lens.

Explanation:

this is right answer