The near point (the smallest distance at which an object can be seen clearly) and the far point (the largest distance at which an object can be seen clearly) are measured for six different people. Near Point(cm) Far Point(cm)

Avishka 40 [infinity]
Berenice 30 300
Chadwick 25 500
Danya 25 [infinity]
Edouard 80 200
Francesca 50 [infinity]

Of the farsighted people, rank them by the power of the lens needed to correct their hyperopic vision. Rank these from largest to smallest power required.

1. Berenice
2. Avishka
3. Francesca
4. Edouard

1. Berenice  = 0.67 D

2. Avishka  = 1.50 D

3. Francesca  = 2.00 D

4. Edouard = 2.75 D

Explanation:

The farsighted people are those with near point greater than 25 cm.

They include Avishka, Berenice, Edouard and Francisca.

A converging lens is needed to correct farsightedness, or hyperopia, therefore, the focal length, f, is positive. The image formed is virtual and on the same side of the lens. Thus the image distance is negative

From the lens formula, 1/f = 1/v  1/u; but v is negative

Therefore, 1/f = 1/u - 1/v

But, power of a lens = 1/f in meters.

Therefore,  P =  1/u - 1/v

For Avishka, u = 25 cm or 0.25 m, v = 0.4 m

P = 1/ 0.25 - 1/0.4 = 1.5 D

For Berenice, u = 0.25 m, v = 0.3 m

P = 1/0.25 - 1/0.30 =0.7 D

For Edouard, u = 0.25 m, v = 0.80 m

P = 1/0.25 - 1/0.80 =2.75 D

For Francesca, u = 0.25 m, v = 0.50 m

P = 1/0.25 - 1/0.5 = 2.0 D

the answer is a)

mark brainliest

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~