The nearpoint of an eye is 123 cm. A corrective lens is to be used to allow this eye to clearly focus on objects 25 cm in front of it.
What should be the focal point of this lens?
B)
What is the power of the needed corrective
lens?
A prescription for a corrective lens calls for +2.50 D. The lensmaker grinds the lens from a “blank” with n=1.56 and a performed convex front surface of radius of curvature of 20.0 cm. What should be the radius of curvature of the other surface
A prescription for a corrective lens calls for +2.50 D. The lensmaker grinds the lens from a “blank” with n=1.56 and a performed convex front surface of radius of curvature of 20.0 cm. What should be the radius of curvature of the other surface?
To find the focal point of the lens that will allow the eye to focus on objects 25 cm in front of it, we can use the lens formula:
1/f = 1/v - 1/u
Where:
f = focal length of the lens
v = image distance (distance where the image is formed)
u = object distance (distance of the object from the lens)
In this case, the object distance (u) is 25 cm, and the nearpoint of the eye is 123 cm. We need to find the focal length (f) of the lens.
Since the eye's nearpoint is the farthest point that the eye can see clearly (when the eye lens is maximally convex), we can consider this as the object distance for the lens. Therefore, u = 123 cm.
Plugging the values into the lens formula, we have:
1/f = 1/25 - 1/123
To simplify the equation, we can find the least common denominator (LCD) of 25 and 123:
LCD(25, 123) = 3075
Multiplying every term in the equation by the LCD, we get:
3075/f = 123 - 25
Simplifying further:
3075/f = 98
To find f, we can isolate it by cross-multiplying:
f * 98 = 3075
Divide both sides by 98:
f = 3075 / 98
Using a calculator, we find that:
f ≈ 31.377 cm
Therefore, the focal point of the lens should be approximately 31.377 cm.