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Andrew [12]
3 years ago
7

A physicist's left eye is myopic (i.e., nearsighted). This eye can see clearly only out to a distance of 35 cm.Find the focal le

ngth and the power of a lens that will correct this myopia when worn 2.0 cm in front of the eye.
Physics
1 answer:
Vikki [24]3 years ago
3 0

Answer:

Explanation:

To get the focal length, we will use the lens formula;

1/f = 1/u + 1/v

f is the focal length

u is the object distance

v is the image distance

Given

since the physicist's left eye is myopic, it will be corrected using concave lens and the image distance is negative.

u = 35cm

v = -2.0

1/f = 1/35-1/2

1/f = 2-35/70

1/f = -33/70

f = -70/33

f = -2.12 cm

f = -0.0212m

Power of a lend is the reciprocal of its focal length

Power of the lens = 1/f

P = 1/-0.0212

P = -47.17dioptres

The power of the lens is -47.17D

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v0x = initial horizontal velocity.

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Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):

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y = -0.881 m

The ball will drop 0.881 m by the time it reaches the catcher.

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