Question

Three converging lenses of focal length \(5.0 \mathrm{~cm}\) are arranged with a spacing of \(2.0 \cdot 10^{1} \mathrm{~cm}\) between them, and are used to image an insect \(2.0 \cdot 10^{1} \mathrm{~cm}\) away. a) Where is the image? b) Is it real or virtual? c) Is it upright or inverted?

Short answer

Based on the given problem involving a series of three converging lenses, with each having a focal length of 5cm, and an insect placed 20cm away from the first lens: a) The final image is 8.18cm away from the third lens. b) The final image is real. c) The final image is inverted.

Step-by-step solution

Use the lens formula for the first lens

To find the position of the image formed by the first lens, we can use the lens formula: \( \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \) where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. Using the values for the first lens \((f=5.0cm, d_o=20cm)\), we can solve for the image distance: \( \frac{1}{5.0} = \frac{1}{20} + \frac{1}{d_i} \)

Calculate the image distance for the first lens

Solve for \(d_i\) and get \(d_i = \frac{1}{\frac{1}{5.0} - \frac{1}{20}}\) \(d_i = 6.67 cm\) The positive image distance means that the image formed by the first lens is real.

Move to the Second Lens and Find Image Position

Now, the image formed by the first lens will act as an object for the second lens, and the distance between this image and the second lens needs to be calculated. Since the spacing between lenses is 20cm and the first image was 6.67cm from the first lens, the distance from the second lens to the first image is 20cm - 6.67cm = 13.33cm. Use the lens formula again but now considering the object distance to the second lens (\(d_o = 13.33cm\)) \( \frac{1}{5.0} = \frac{1}{13.33} + \frac{1}{d_i} \)

Calculate the image distance for the second lens

Solve for \(d_i\) and get \(d_i = \frac{1}{\frac{1}{5.0} - \frac{1}{13.33}}\) \(d_i = -16.67 cm\) A negative result for image distance indicates that the image formed by the second lens is virtual.

Move to the Third Lens and Find Image Position

The distance between the virtual image created by the second lens and the third lens can be calculated by adding the distances between the lenses and the image formed: 20cm + 16.67cm = 36.67cm. Thus, for the third lens, \(d_o = 36.67cm\). Use the lens formula for the third lens: \( \frac{1}{5.0} = \frac{1}{36.67} + \frac{1}{d_i} \)

Calculate the image distance for the third lens

Solve for \(d_i\) and get \(d_i = \frac{1}{\frac{1}{5.0} - \frac{1}{36.67}}\) \(d_i = 8.18 cm\) A positive image distance means that the image formed by the third lens is real. a) The final image is 8.18cm away from the third lens. b) The final image is real (since the image distance is positive). c) To find the orientation of the images, we keep track of any flipping of the images as the lenses are passed: Initially (first lens), the image will be inverted, since it's a real image. For the second lens, the image has a virtual image with positive magnification (due to the negative image distance), which means it flips the image orientation. The new orientation labeled as "upright." Finally, for the third lens, the image is real, so the orientation will be inverted.