- Power of the eye is 62.5 diopter
- The size of the image is - 0.5 mm
- The size of the image be if the object were held at the normal 25.0 cm is -0.16 mm
What is the power of the eye when viewing an object?
- The power of the eye varies from 50.0 D (for distant entirely relaxed vision) to 54.0 D (for near fully accommodated vision), which is an increase of 8%, for an eye with this usual 2.00 cm lens-to-retina distance.
- In other words, multiply 50.5 diopters by 1 over 2.00 times 10 to the minus 2 meters, then take the difference and multiply the result by 1 then multiply that result by the exponent to obtain 2.00 meters.
- As a result, they can only see a maximum of 2 meters in front of them without glasses since they are severely nearsighted.
- It is 2.00 cm from the lens to the retina.
Given details :
a) For this exercise we use the equation for thin lenses
1 / f = 1 / p + 1 / q
where f is the focal length, p the distance to the object and q the distance to the image
- In this case q is the distance from the lens to the retina q = 2.00 cm
- The distance p is the distance from the object to the lens p = 8.00 cm
1 / f = 1 / 8.00 + 1/2
1 / f = 0.625 cm⁻¹
power is the inverse of the focal length in meters
P = 1 / f [m]
P = 0.625
P = 62.5 diopter
b) the size of the image
m = - p / q
m = - 2 / 8.00
m = -0.25 X
if we assume that the object is right
m = h´ / h
h´ = m h
h´ = -0.25 x 2
h´ = -0.5 mm
if the height of the object h is straight the image and inverted
c) if the object were at p = 25.0 cm
m = - 2/25
m = -0.08
h´ = -0.08 x 2
h´ = -0.16 mm
the image would be much smaller
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