A 4.5 cm long insect is perpendicular to the optical axis of a thin lens; the bug is 5.4 cm away from the lens. The lens forms an inverted image of the insect which is 0.9 cm long on a screen behind the lens.
a. How far behind the lens is the screen?
b. What is the lens' focal length (pay attention to sign)?
cm
c. If the insect were instead only 0.25 cm from the lens, what would the image distance be (pay attention to sign)?
cm
d. Could the new image be projected on a screen?
To solve this problem, we can use the thin lens equation:
1/f = 1/d₀ + 1/d_i
Where:
f is the focal length of the lens,
d₀ is the object distance (distance from the object to the lens), and
d_i is the image distance (distance from the lens to the image).
a. To find the distance behind the lens where the screen is located, we need to calculate the image distance (d_i) using the thin lens equation. We know that the insect is 5.4 cm away from the lens (d₀ = 5.4 cm), and the image of the insect is formed on the screen and is 0.9 cm long.
Since the image is inverted, the image distance (d_i) will be negative:
1/f = 1/d₀ + 1/d_i
1/d_i = 1/f - 1/d₀
Now substitute the known values:
1/d_i = 1/f - 1/d₀
1/d_i = 1/f - 1/5.4 cm
1/d_i = (5.4 - f)/(5.4 * f) [Multiplying both sides by f * 5.4]
Since the image distance (d_i) is negative, we can determine it as:
d_i = - (5.4 * f) / (5.4 - f) [Inverting both sides]
Now, substitute the length of the insect (0.9 cm) and solve for d_i:
0.9 cm = - (5.4 cm * f) / (5.4 cm - f)
0.9 cm * (5.4 - f) = - (5.4 cm * f)
4.86 cm - 0.9f cm = -5.4f cm
4.86 cm = -5.4f cm + 0.9f cm
4.86 cm = -4.5f cm
Now, isolate f to find the focal length:
4.86 cm = -4.5f cm
f = -4.86 cm / -4.5 cm
Therefore, the focal length f = 1.08 cm (rounded to two decimal places).
b. The lens' focal length is 1.08 cm.
c. If the insect were instead only 0.25 cm from the lens (d₀ = 0.25 cm), we can use the same thin lens equation to find the new image distance (d_i):
1/d_i = 1/f - 1/d₀
1/d_i = 1/1.08 cm - 1/0.25 cm
Simplifying:
1/d_i = (0.926 cm - 4 cm) / (1.08 cm * 0.25 cm)
1/d_i = -3.074 / 0.27 cm
d_i ≈ -11.38 cm (rounded to two decimal places)
Therefore, if the insect were 0.25 cm from the lens, the image distance (d_i) would be approximately -11.38 cm.
d. The new image distance is negative, which means the image is formed on the same side as the object (in front of the lens). Since the image distance is negative and the screen is behind the lens, the new image formed cannot be projected on a screen.