An eye that can see far away object clearly but which can’ t clearly focus on objects that are 25 cm from the lens of the eye is farsighted or hyperoptic. The near point for such eye is farther than 25 cm from the eye. Correction is done using a converging lens that images an object at the normal near point to the person’s larger than normal
near point.
An eye that can see objects at a normal near point of 25 cm but has a far point that is smaller than very far away at infinity is nearsighted or myopic. The far point of such an
eye is closer than infinity. Correction is done using a diverging lens that images an object that is very far away to the persons small than normal far point.
To do
Use the application to investigate a normal eye for an object from the normal near point to the far point.
Use the application to investigate a farsighted eye for an object from the normal near point to the far point.
Explain how the corrective lens works for a farsighted eye.
Use the application to investigate a nearsighted eye for an object from the normal near point to the far point.
Explain how the corrective lens works for near sighted eye..
Use Excel to set up calculations for a corrective lens of a person’s eye. Inputs are the near point distance, the far point distance and the distance of the lens from the eye
to be corrected. Outputs are the power in diopters of the required corrective lens. Plan on making two sheets in the Excel file – one sheet for a farsighted eye, and
another sheet for a near sighted eye.
Assuming an index of refraction of 1.42 for the lens, draw a one – to – one lens that.
Has a power of 3.4D.
Has a power of – 2.5 D.

a. When glasses (or contact lenses) are used to correct nearsightedness, the lens should form the image at the far point.

b. If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 the contact lenses that the person would need to see an object at infinity clearly?

f 1 = -3.50 m

Explanation:

a. Nearsightedness (the person cannot focus on the far point objects) physics of eyesight correction,  can be resolved by using focal length of concave lens as the corrective lens, which correct the eyesight artificially, as the eye’s lens bends the rays from a distant object too much and the rays are brought to a focus before they reach the retina.  A diverging lens bends the ray outwards before they reach the eye’s lens and the rays are brought to a focus on the retina.  

The lens formula to find the focal length of the lens is:

                      1/f = 1u + 1/v

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

The light lands in front of the retina, so the image is forming in front of the retina .When concave lens used as a corrective lens the clear vision is achieved, helping to bend the rays of light outwards and the light converge further and reach the retina to form the imagemeaning the lens should form the image at the far point.

b.      The lens formula is,

                       1/f = 1u + 1/v

Here the object distance is infinity,so the reciprocal of infinity gives zero.

   Substitute 0.0  for u   and 0.0 -3.50 m    for   v  

1/f = 1/ -3.50m  + 1/ 0.0

Therefore, the focal length of the contact lenses that the person would need to see an object clearly is -3.50 m

the answer to your question is b

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