AP Physics 2 Exam Question:
A concave lens with a focal length of -15 cm is used to correct nearsightedness. A student with a nearsightedness of -2.5 diopters is prescribed this lens. The student wants to know the position, size, and nature of the image formed by this lens when an object is placed 30 cm away from it. Assuming the lens is thin and the index of refraction of the lens is 1.5, calculate the position, size, and nature of the image formed.
Answer:
Given data:
First, let's convert the nearsightedness from diopters to meter^(-1) or 1/m by using the formula:
s = 1/f
where s is the nearsightedness in diopters and f is the focal length in meters.
Since f is given in centimeters, we convert it to meters by dividing by 100:
f = -15 cm ÷ 100 = -0.15 m
Substituting the given nearsightedness in diopters:
-2.5 = 1/f
Solving for f:
f = -1/2.5 = -0.4 m^(-1)
Now, let's calculate the object distance (u) from the lens, using the lens maker's formula:
1/f = (n-1)(1/r1 - 1/r2)
where r1 and r2 are the radii of curvature of the lens' surfaces.
Since the lens is assumed to be thin, we can consider its two surfaces to be nearly parallel, so the radii of curvature are both infinite, making 1/r1 and 1/r2 equal to zero.
1/f = (n-1)(1/0 - 1/0) = (n-1)(0 - 0) = 0
Since the object distance (u) is positive, we take it as the absolute value of the given object distance:
u = |do| = |30 cm| = 30 cm = 0.3 m
Now, using the lens formula:
1/f = 1/v - 1/u
Solving for the image distance (v):
1/v = 1/f + 1/u 1/v = 1/-0.4 + 1/0.3 1/v = -2.5 + 3.33 1/v = 0.83
v = 1/(0.83) v = 1.20 m
The result is positive since the image formed is on the same side as the object (the image distance is measured from the lens).
Next, let's calculate the magnification (m) of the image:
m = -v/u m = -1.20/0.3 m = -4
The negative magnification indicates an inverted image.
Finally, let's determine the size of the image (h') relative to the size of the object (h):
m = h'/h
Solving for h':
h' = m * h h' = -4 * h
Given that the size of the object is not specified, we can assume it to be 1 unit (arbitrary).
h' = -4 * 1 h' = -4 units
The size of the image is 4 units, and since the magnitude is greater than 1, it indicates that the image is larger than the object.
In conclusion, the image formed by the concave lens is located 1.20 meters away from the lens, is 4 units in size (larger than the object), and is inverted in nature.