How to Analyze a Two-Lens System with Converging and Diverging Lenses

TL;DR

To analyze a two-lens system, apply the thin lens equation for each lens: use a positive focal length for the converging lens and a negative focal length for the diverging lens. Calculate the image position, nature, orientation, and size based on the object's distance from each lens, noting that a converging lens can produce real images while a diverging lens always produces virtual images.

Transcript

in this video we're going to go over a two- lens system a system with a converging lens and a Divergent lens the equation that we need is this equation f is the focal length D is the distance between the object and the center of the lens and Di is the distance between the image and the center of the lens magnification is equal to di D and it's also... Read More

Key Insights

• ❎ The focal length of a converging lens is positive, while the focal length of a diverging lens is negative.
• 🤫 A negative focal length indicates a diverging lens, which always produces a virtual image.
• 🧘 The distance between the object and the lens determines the position of the image, with positive values indicating a real image and negative values indicating a virtual image.
• ❓ The magnification value determines the size of the image relative to the object, with values greater than one indicating enlargement and values less than one indicating reduction.
• ❓ The orientation of the image depends on whether it is upright or inverted, which is determined by the magnification value.
• 🧘 The combination of a converging lens and a diverging lens can produce both real and virtual images, depending on the position of the object and the lens.
• 🫥 The positions and sizes of the images can be verified using ray diagrams, where converging lines represent real images and dashed lines represent virtual images.

Questions & Answers

Q: How do you solve a two-lens system with a converging and a diverging lens?

Apply the thin lens equation 1/f = 1/do + 1/di to each lens in turn. Solve the first (converging) lens for its image distance di, then treat that image as the object for the second (diverging) lens, using the spacing between the lenses to find the new object distance. In the example the lenses are 18 cm apart, so the first image at di = 6 cm gives an object distance of 18 - 6 = 12 cm for the second lens.

Q: Is a converging lens concave or convex?

A converging lens is convex. It is thicker at the center and brings light rays together. A diverging lens is the opposite: it is concave, thinner at the center, and spreads light rays out.

Q: Should the focal length be positive or negative for each lens?

For a converging (convex) lens the focal length is positive, and for a diverging (concave) lens it is negative. In the example the converging lens uses f = +3 cm while the diverging lens uses f = -4 cm when plugged into the thin lens equation.

Q: What image will a divergent lens produce here?

The diverging lens produces a virtual image. With an object distance of 12 cm and f = -4 cm, the equation gives di = -3 cm. Because di is negative the image is virtual and forms on the left side of the diverging lens, between the lens and its focal point (f = 4 cm).

Q: How do you tell whether an image is real or virtual?

Check the sign of the image distance di. When di is positive the image is real and forms on the right side of the lens; when di is negative the image is virtual and forms on the left side. The first lens gives di = +6 cm (real), and the second gives di = -3 cm (virtual).

Q: What does the magnification tell you about the image?

Magnification equals -di/do and also equals the ratio of image height to object height. A negative value means the image is inverted and a positive value means it is upright. If the absolute value is greater than one the image is enlarged, less than one it is reduced, and exactly one means it is the same size as the object.

Q: Why is the first image both inverted and the same size as the object?

For the converging lens do = 6 cm and di = 6 cm, so the magnification is -6/6 = -1. The negative sign makes the image inverted, while the magnitude of 1 means the image height equals the object height. This happens because the object sits at twice the focal length (2f), where do and di are equal.

Q: How can you verify the image positions and sizes?

You can confirm the calculated positions and sizes with ray diagrams. Converging (solid) lines that actually meet represent a real image, while dashed lines traced backward represent a virtual image. This matches the real first image on the right of the converging lens and the virtual second image on the left of the diverging lens.

Summary & Key Takeaways

• The video discusses the equations and principles involved in a two-lens system, including the thin lens equation and magnification.

• It provides an example scenario with a converging lens and a diverging lens, where the distances and heights of the object and image are given.

• The video explains how to calculate the position, nature (real or virtual), orientation (upright or inverted), and size (enlarged or reduced) of the resulting images from each lens.