Two systems are formed from a converging lens and a...

Two systems are formed from a converging lens and a diverging lens, as shown in parts a and b of the drawing. (The point labeled $" F_{\text { comverging }} "$ is the focal point of the converging lens.) An object is placed inside the focal point of lens 1 at a distance of 10.00 cm to the left of lens 1. The focal lengths of the converging and diverging lenses are 15.00 and 20.0 cm, respectively. The distance between the lenses is 50.0 cm. Determine the final image distance for each system, measured with respect to lens 2.

Key Concepts

Lens Formula

The lens formula, given by 1/f = 1/d_o + 1/d_i, quantitatively relates the focal length (f) of a lens to the distance of the object (d_o) and the image (d_i). This equation is fundamental for determining the position, size, and nature (real or virtual) of an image produced by a thin lens.

Converging Lens

A converging lens, typically a convex lens, bends parallel incoming light rays so that they meet at a single point known as the focal point. Depending on the object's location relative to the focal point, the lens can form either real and inverted images or virtual and upright images.

Diverging Lens

A diverging lens, usually a concave lens, spreads parallel light rays apart so that they appear to come from a common focal point on the same side of the lens as the object. This type of lens typically forms virtual, upright, and diminished images regardless of the object's position.

Compound Lens System

A compound lens system comprises two or more lenses arranged in sequence such that the image produced by one lens serves as the object for the subsequent lens. Analyzing such systems involves applying the lens formula sequentially while carefully accounting for the distances between the lenses and the resultant image positions.

Optical Sign Conventions

Optical sign conventions provide a consistent framework for assigning positive or negative values to quantities such as object distance, image distance, and focal length. These conventions help distinguish between real and virtual images and are crucial for correctly applying the lens formula in complex optical systems.

Transcript

00:01 So for part a, we have this system where there's a lens l1, and the focal length for l1 is equal to 15 centimeters.

00:13 And then there's another lens for which the focal length is negative 20 centimeters because that's a diverging lens.

00:20 And the two lenses are a distance 50 centimeter apart.

00:25 And there is an object such that do is equal to 10 centimeters from the first lens.

00:34 And we have to find the location of the final image.

00:39 Now, when there's a combination of lenses, what we would do is we would find the image, location of the image for the first lens, and then use that as the object for the second lens.

00:49 So we apply the thin lens formula, 1 over do plus 1 over d .i, is equal to 1 over f, and that would be 1 over 10 plus 1 over d .i is equal to 1 over 15, or 1 over di is equal to 1 over 15 minus 1 over 10, and solving that for d .i gives d .i is equal to negative 30 centimeters.

01:18 So the image is formed a distance of 30 centimeters from the lens on the left.

01:24 So that means now that the distance from the second lens, let's call the do prime, that would be equal to 50 minus negative 30 centimeters, which is equal to 80 centimeters.

01:39 So we can now use this in the same thin lens formula.

01:43 So we would say 1 over 80 plus 1 over d .i is equal to negative 1 over 20.

01:51 Now, f is negative because the lens is a diverging lens.

01:56 So we have 1 over d .i.

01:59 Is equal to negative 1 over 20 minus 1 over 80.

02:03 And solving that for d .i gives d .i is equal to negative 16 centimeters.

02:09 So what that means is that the image is 16 .0 centimeters to the left of...

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