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?
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
To determine the focal length lenses needed to correct the person's vision, you need to consider the relationship between the object distance (the near point) and the image distance.
In this case, the person's near point is 60.0 cm from his eyes instead of the usual 25.0 cm. The object distance is the distance from the lens to the near point, which is 60.0 cm + 2.0 cm (the distance the lens is worn in front of the eye), resulting in a total object distance of 62.0 cm.
To find the focal length of the lenses, you can use the lens formula:
1/f = 1/di - 1/do
Where:
f is the focal length of the lens
di is the image distance
do is the object distance
Since the glasses are worn in front of the eye, the image distance (di) will be negative because the image is formed on the same side as the object. The image distance can be considered equal to the distance from the lens to the eye, which is 2.0 cm.
Substituting the given values into the lens formula, we have:
1/f = 1/-2.0 cm - 1/62.0 cm
Simplifying this equation, we get:
1/f = -31/62 - 1/62
1/f = -32/62
Inverting both sides of the equation, we find:
f = -62/32
f ≈ -1.9375 cm
The negative sign indicates that the lenses should be concave (diverging) lenses since the person's near point is farther than usual.
Therefore, the focal length lenses needed to correct the person's vision are approximately 1.9375 cm.