Two converging lenses with the focal lengths of 40 cm and 20 cm are 10 cm apart. a 2.0-cm-tall object is 15 cm in front of the 40-cm-focal-length lens. Calculate the image position and height.
To solve this problem, we will first find the image formed by the first lens (40 cm focal length) and then treat this image as the object for the second lens (20 cm focal length) to find the final image position and height.
First, we will use the lens equation for the first lens:
1/f = 1/do + 1/di,
where f is the focal length, do is the object distance, and di is the image distance.
For the first lens, f = 40 cm and do = 15 cm. Plug these values into the equation:
1/40 = 1/15 + 1/di
Now solve for di:
1/di = 1/40 - 1/15 = -1/60
di = -60 cm
The negative sign indicates that the image is an inverted virtual image, which is formed on the same side as the object. The distance between the first image and the first lens is 60 - 15 = 45 cm.
Now we will find the image height:
Image height = (Image distance / Object distance) × Object height
Image height = (45 / 15) × 2.0 = 6.0 cm
Now we will treat this image as an object for the second lens (20 cm focal length). For the second lens, the new object distance is:
do' = 45 cm - 10 cm = 35 cm (since the lenses are 10 cm apart)
Now use the lens equation for the second lens:
1/f' = 1/do' + 1/di',
where f' is the focal length of the second lens, do' is the new object distance, and di' is the final image distance.
For the second lens, f' = 20 cm and do' = 35 cm. Plug these values into the equation:
1/20 = 1/35 + 1/di'
Now solve for di':
1/di' = 1/20 - 1/35 = 3/140
di' = 140/3 = 46.67 cm
Next, we will find the magnification of the second lens:
Magnification = Image distance / Object distance
Magnification = 46.67 / 35 = 1.333
Now find the final image height by multiplying the height of the image formed by the first lens by the magnification of the second lens:
Final image height = Magnification × Image height
Final image height = 1.333 × 6 = 8.0 cm
So, the final image is formed at a distance of 46.67 cm on the other side of the second lens and has a height of 8.0 cm.
To solve this problem, we can use the lens formula:
\(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\)
Where:
- \(f\) is the focal length of the lens
- \(v\) is the image distance from the lens
- \(u\) is the object distance from the lens
Given:
- Focal length of the first lens (\(f_1\)) = 40 cm
- Focal length of the second lens (\(f_2\)) = 20 cm
- Distance between the lenses = 10 cm
- Object distance in front of the first lens (\(u_1\)) = 15 cm
- Object height (\(h\)) = 2.0 cm
We can divide the problem into three stages:
1. First lens (Lens 1)
2. Space between the lenses
3. Second lens (Lens 2)
Stage 1: First Lens (Lens 1)
Using the lens formula, we can find the image distance from Lens 1 (\(v_1\)):
\(\frac{1}{f_1} = \frac{1}{v_1} - \frac{1}{u_1}\)
Substituting the given values:
\(\frac{1}{40} = \frac{1}{v_1} - \frac{1}{15}\)
Rearranging the equation:
\(\frac{1}{v_1} = \frac{1}{40} + \frac{1}{15}\)
\(\frac{1}{v_1} = \frac{15 + 40}{600}\)
\(\frac{1}{v_1} = \frac{55}{600}\)
Taking the reciprocal, we find:
\(v_1 = \frac{600}{55}\)
\(v_1 \approx 10.91\) cm
Stage 2: Space Between the Lenses
The image formed by the first lens (Lens 1) will act as an object for the second lens (Lens 2). So, the distance between the lenses (\(d\)) will be equal to the image distance from the first lens (\(v_1\)).
Thus, \(d = v_1\)
\(d \approx 10.91\) cm
Stage 3: Second Lens (Lens 2)
The object distance in front of Lens 2 (\(u_2\)) will be equal to the distance between the lenses (\(d\)) minus the focal length of Lens 1 (\(f_1\)).
Thus, \(u_2 = d - f_1\)
\(u_2 = 10.91 - 40\)
\(u_2 \approx -29.09\) cm (negative sign indicates object is on the same side as the second lens)
Using the lens formula, we can find the image distance from Lens 2 (\(v_2\)):
\(\frac{1}{f_2} = \frac{1}{v_2} - \frac{1}{u_2}\)
Substituting the given values:
\(\frac{1}{20} = \frac{1}{v_2} + \frac{1}{29.09}\)
Rearranging the equation:
\(\frac{1}{v_2} = \frac{1}{20} - \frac{1}{29.09}\)
\(\frac{1}{v_2} = \frac{29.09 - 20}{20 \times 29.09}\)
\(\frac{1}{v_2} = \frac{9.09}{581.8}\)
Taking the reciprocal, we find:
\(v_2 = \frac{581.8}{9.09}\)
\(v_2 \approx 63.89\) cm
Now, we can calculate the magnification (\(m\)) of the second lens (Lens 2) using the formula:
\(m = \frac{h_2}{h_1} = -\frac{v_2}{u_2}\)
Substituting the given values:
\(m = -\frac{63.89}{-29.09}\)
\(m \approx 2.198\)
To find the height of the image formed by Lens 2 (\(h_2\)), we multiply the magnification (\(m\)) by the height of the object (\(h_1\)):
\(h_2 = m \times h_1\)
\(h_2 = 2.198 \times 2.0\)
\(h_2 \approx 4.396\) cm
Therefore, the image position is approximately 63.89 cm from the second lens (Lens 2), and the height of the image is approximately 4.396 cm.
To calculate the image position and height, you can use the lens formula and magnification formula.
The lens formula relates the object distance (u), the image distance (v), and the focal length (f) of a lens:
1/f = 1/v - 1/u
The magnification formula relates the object height (h) and image height (h') to the object distance (u) and image distance (v):
m = -v/u = h'/h
Here's how you can solve the problem step by step:
Step 1: Identify the given values:
- The focal length of the first lens (f1) is 40 cm.
- The focal length of the second lens (f2) is 20 cm.
- The distance between the lenses is 10 cm.
- The object height (h) is 2.0 cm.
- The object distance (u) is 15 cm.
Step 2: Calculate the image position (v1) using the lens formula for the first lens:
1/f1 = 1/v1 - 1/u
1/40 = 1/v1 - 1/15
Rearranging the equation:
1/v1 = 1/40 + 1/15
1/v1 = (15 + 40)/(15 * 40)
1/v1 = 55/600
Taking the reciprocal of both sides:
v1 = 600/55
v1 ≈ 10.91 cm
Step 3: Calculate the distance between the first lens and the second lens:
The total distance between the first lens and the object:
d1 = u - v1
d1 = 15 - 10.91
d1 ≈ 4.09 cm
The distance between the lenses minus d1 gives the distance between the second lens and the image:
d2 = 10 - d1
d2 ≈ 10 - 4.09
d2 ≈ 5.91 cm
Step 4: Calculate the image position (v2) using the lens formula for the second lens:
1/f2 = 1/v2 - 1/u
1/20 = 1/v2 - 1/d2
Rearranging the equation:
1/v2 = 1/20 + 1/d2
1/v2 = (d2 + 20)/(20 * d2)
1/v2 = (5.91 + 20)/(20 * 5.91)
Taking the reciprocal of both sides:
v2 = (20 * 5.91)/(5.91 + 20)
v2 ≈ 15.44 cm
Step 5: Calculate the image height (h') using the magnification formula for the second lens:
m = -v/u = h'/h
Rearranging the equation:
h' = m * h
h' = (-v2/u) * h
h' = (-15.44/15) * 2.0
h' ≈ -2.05 cm
The negative sign indicates that the image is inverted.
Therefore, the image position is approximately 15.44 cm from the second lens, and the image height is approximately -2.05 cm.