(a) A 2.20 cm high insect is 1.24 m from a 135 mm focal-length lens. What is the image distance?
mm
How high is the image?
cm What type is it? (Select all that apply.)
The image is upright.
The image is inverted.
The image is behind the lens.
The image is in front of the lens.
The image is virtual.
The image is real.
(b) What if f = -135 mm? What is the image distance?
mm
How high is the image?
cm What type is it?
The image is behind the lens.
The image is real.
The image is virtual.
The image is upright.
The image is inverted.
The image is in front of the lens.
To solve both parts of the question, we 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, and do is the object distance.
(a) Given:
Height of the insect (ho) = 2.20 cm
Object distance (do) = 1.24 m = 12400 mm
Focal length of the lens (f) = 135 mm
To find the image distance, we rearrange the lens formula:
1/di = 1/f - 1/do
Substituting the known values:
1/di = 1/135 - 1/12400
Calculating this expression, we get:
1/di ≈ 0.00741 - 0.00008
1/di ≈ 0.00733
Taking the reciprocal of both sides:
di ≈ 136 mm
Therefore, the image distance is approximately 136 mm.
To find the height of the image, we can use the magnification formula:
magnification (m) = -di/do
Substituting the known values:
m ≈ -(136 mm) / (12400 mm) ≈ -0.01097
The negative sign indicates that the image is inverted.
The height of the image (hi) can be found using the magnification formula:
magnification (m) = hi/ho
Substituting the known values:
-0.01097 = hi / 2.20
Solving for hi:
hi ≈ -0.01097 * 2.20 ≈ -0.02414 cm
Since the height of the image is negative, it indicates that the image is inverted.
Therefore, the image distance is approximately 136 mm, and the height of the image is approximately 0.02414 cm. The image is inverted.
(b) Given:
Focal length of the lens (f) = -135 mm
Using the same calculations as in part (a):
1/di = 1/f - 1/do
1/di = 1/(-135) - 1/12400
1/di ≈ -0.00741 - 0.00008
1/di ≈ -0.00749
Taking the reciprocal of both sides:
di ≈ -134 mm
Therefore, the image distance is approximately -134 mm. The negative sign indicates that the image is behind the lens.
To find the height of the image, we use the magnification formula:
m = -di/do
m ≈ -(-134 mm) / (12400 mm) ≈ 0.01081
The positive sign indicates that the image is upright.
The height of the image can be found using the magnification formula:
m = hi/ho
0.01081 = hi / 2.20
Solving for hi:
hi ≈ 0.01081 * 2.20 ≈ 0.02378 cm
Therefore, the image distance is approximately -134 mm, and the height of the image is approximately 0.02378 cm. The image is upright and behind the lens.