To calculate the power of the lens required for distinct vision, we can use the lens formula:

1/f = 1/v - 1/u

Where:
- f is the focal length of the lens
- v is the image distance (distance of the object from the lens)
- u is the object distance (distance of the image from the lens)

In this case, the person can see objects clearly only beyond 100 cm. This means that the image distance (v) is 100 cm, and the object distance (u) is the distance between the eye and the lens.

We can assume that the distance between the eye and the lens is the least distance of distinct vision, which is typically considered to be 25 cm (0.25 m). However, in this case, we need to consider the distance beyond which the person can see clearly, which is 100 cm (1 m). Therefore, the object distance (u) is 1 m.

Now, let's substitute the values into the lens formula and solve for the focal length (f):

1/f = 1/1 - 1/0.25
1/f = 4 - 1
1/f = 3

Therefore, the focal length (f) is 1/3 meters or 0.33 meters.

To calculate the power of the lens, we can use the formula:

P = 1/f

Substituting the value of the focal length (f), we get:

P = 1/0.33
P ≈ 3.03 diopters

So, the power of the lens required for distinct vision is approximately +3.03 diopters.

In summary:
- Object distance (u) = 1 m
- Image distance (v) = 1 m (100 cm)
- Focal length (f) = 0.33 m
- Power of the lens (P) ≈ +3.03 diopters

This person's near point is actually receding away....so this person is actually suffering from hypermetropia. For that we need to use a convex lens for correction.... the power of the lens required is as follows :-u = -25 cm ( normal near point )v = -100 cm (near point of this defective eye)f= ? we have,1/v - 1/u = 1/f (lens formula)=> (1/-100) - (1/-25) = 1/f=> -1+4/100 = 1/f=> 3f = 100=> f = 33.3 cm=> f = 0.33 mso,P = 1/f = 1/0.33 = 3.0 mhence, the power of the lens required is + 3.0 mhope this helped :-)?