The near and the far points of a short-sighted are 8 cm and 20 cm from the eyeball.

a) What power of corrective lens is needed for this eye to see distant objects clearly?
b)What will the resulting near point then be?(Assume that the lens is placed very close to the eye)

Answers so far:
a) 1/f = 1/u + 1/v
1/f = 1/0.08 + 1/0.20
1/f = 17.5
P = 1/f
P = 17.5 D

b) 1/∞ + 1/-0.2 = 1/f
f = 0

To find the power of the corrective lens needed for the short-sighted eye to see distant objects clearly, we can use the lens equation:

1/f = 1/u + 1/v

Where f is the focal length of the lens, u is the object distance, and v is the image distance.

a) For the short-sighted eye, the near point is given as 8 cm, so the object distance (u) is 8 cm. The far point is given as 20 cm, so the image distance (v) would be -20 cm (since the image is formed on the same side as the object for a short-sighted eye). Plugging these values into the lens equation:

1/f = 1/0.08 + 1/(-0.20)
1/f = 12.5 - 5
1/f = 7.5

To find the power of the lens (P), we use the equation:

P = 1/f

P = 1/7.5
P = 0.13 D (rounded to two decimal places)

Therefore, the power of the corrective lens needed for this eye to see distant objects clearly is approximately 0.13 D.

b) To find the resulting near point when the corrective lens is placed very close to the eye, we need to use the lens equation again. When the lens is placed very close to the eye, the object distance (u) can be considered as positive infinity (∞). Plugging this value into the lens equation:

1/∞ + 1/(-0.20) = 1/f

Since 1/∞ is essentially zero, the equation simplifies to:

0 + 1/(-0.20) = 1/f

Solving for f:

1/(-0.20) = 1/f

-5 = 1/f

f = -1/5
f = -0.20 m

Since the near point is the object distance for a nearsighted eye, the resulting near point when the corrective lens is placed very close to the eye would be approximately -0.20 m.

Note: The negative sign indicates that the near point is closer to the eye than the lens.