Ordinary glasses are worn in front of the eye and usually 2.00 cm in front of the eyeball. A certain person can see distant objects well, but his near point is 60.0 cm from his eyes instead of the usual 25.0 cm. Suppose that this person needs ordinary glasses.
What focal length lenses are needed to correct his vision ?
Here's my work so far:
I think that since the near point is 60cm, the image should be at 60cm from the lens, but since the lens is worn 2cm from the eye, the focal length should be 60-2 so 58cm, right?
That's the wrong answer. The mastering physics program didn't accept it.
To find the focal length lenses needed to correct the vision of the person, you need to use the lens equation, which relates the distance of the object (do), the distance of the image (di), and the focal length (f) of the lens. The lens equation is given by:
1/f = 1/do + 1/di
In this case, the object distance (do) is 60 cm (the near point), and the image distance (di) should be 2 cm in front of the eye since the glasses are worn 2 cm from the eye.
Substituting the values into the lens equation:
1/f = 1/60 + 1/-2
Making sure to use the appropriate signs for the distances, where the object distance is positive and the image distance is negative since it is on the opposite side of the lens.
Simplifying the equation:
1/f = 1/60 - 1/2
1/f = (1 - 30)/60
1/f = 29/60
To find the focal length, take the reciprocal of both sides:
f = 60/29
Therefore, the focal length of the lenses needed to correct the person's vision is approximately 2.07 cm.
Actually, your calculation is not correct. The near point is the closest distance at which a person can see clearly, and in this case, it is given as 60.0 cm. To correct the person's vision, we need to find the focal length of the lenses that will allow the person to see clearly at this near point.
The formula to calculate the focal length of lenses is:
1/f = 1/v - 1/u
Where:
f is the focal length
v is the image distance (distance of the image from the lens)
u is the object distance (distance of the near point from the lens)
Given that the near point is 60.0 cm and the lens is worn 2.00 cm from the eyeball, the object distance (u) would be 60.0 cm + 2.00 cm = 62.0 cm.
We can assume that the image distance (v) is the distance at which the lens is worn, which is 2.00 cm.
Plugging these values into the formula, we have:
1/f = 1/2.00 cm - 1/62.0 cm
Simplifying this equation gives:
1/f = (31.0 - 1) / (62.0 * 2.00)
1/f = 30.0 / 124.0
f = 124.0 cm / 30.0
f ≈ 4.13 cm
Therefore, the focal length of the lenses that the person needs to correct their vision is approximately 4.13 cm.
A converging or positive lens is required, with
a power such that when an object is placed 25 cm in front of the
lens, the lens forms a virtual image.
The image distance di= 60 cm, so that this image serves as an
object for the eye at the eye’s near point.
Taking into account the distance 2 cm
di =0.62 m, do =0.27m
Applying the thin lens equation, we find
1/F =1/do +1/di = 1/0.27 +1/0.62 =3.7+1.61 =5.31.
F = 0.188 m