An object is placed 15.0 cm from a first converging lens of...

An object is placed 15.0 cm from a first converging lens of focal length 10.0 cm. A second converging lens with focal length 5.00 cm is placed 10.0 cm to the right of the first converging lens. (a) Find the position $q_{1}$ of the image formed by the first converging lens. (b) How far from the second lens is the image of the first lens? (c) What is the value of $p_{2},$ the object position for the second lens? (d) Find the position $q_{2}$ of the image formed by the second lens. (e) Calculate the magnification of the first lens. (f) Calculate the magnification of the second lens. (g) What is the total magnification for the system? (h) Is the final image real or virtual? Is it upright or inverted (compared to the original object for the lens system)?

Key Concepts

Lens Equation

The lens equation, formulated as 1/f = 1/p + 1/q (where f is the focal length, p is the object distance, and q is the image distance), is fundamental in optics for relating these distances. It allows one to determine the position of an image given the object distance and the lens's focal length, or vice versa.

Focal Length

The focal length of a lens is a measure of how strongly the lens converges or diverges light. It is a key property that determines the behavior of a lens in forming images, and it is essential for calculating image distance and magnification using the lens equation.

Object and Image Distance

Object distance (p) and image distance (q) are the distances from the lens to the object and the image, respectively. Knowing these distances is crucial in applying the lens equation correctly and understanding where images are formed in relation to the lens, whether they are real or virtual.

Magnification

Magnification in optics is the factor by which the size of the image is enlarged or reduced relative to the object, and it is given by the ratio q/p in the context of a lens. This concept not only tells us about the image size but also indicates whether the image is inverted or upright compared to the original object.

Sign Convention in Optics

The sign convention in optics provides a consistent method for assigning positive and negative values to quantities such as object distance, image distance, and focal length. It is essential for accurately determining the nature (real or virtual) and orientation (upright or inverted) of images formed by lenses.

Sequential Optical Systems

In systems involving multiple lenses, the image formed by the first lens typically serves as the object for the next lens. Understanding sequential optical systems is critical because the final image is determined by the combined effect of all lenses, and one must carefully analyze each stage individually while considering the inter-lens distances.

Transcript

00:01 For this question, we have two converging lenses.

00:04 The focal length of the first one, f1, is 10 centimeters.

00:07 And the focal length of the second one, f2, is 5 centimeters.

00:10 And they are a distance d, 10 centimeters apart, f2 being to the right of f1.

00:16 And we have an object that's placed 15 centimeters from the first converging lens.

00:21 So therefore, p1, or the object for the first focal lens, is 15 centimeters.

00:27 And it's positive since it's on the left side.

00:29 And for part a, we're asked to find the image produced by the first lens.

00:38 So we're going to use 1 over p plus 1 over q equals 1 over f, q being the image distance from the lens.

00:50 So we can solve this equation for q.

00:53 And in this case, it's q1 because we're considering the first lens.

00:56 Q1 would be p1 times f1 divided by p1 minus f1 okay and this comes out to equal 30 centimeters it's positive so that means it's to the right of that first lens part b asked to figure out how far from the second lens is the image of the first lens okay and let's box this in as our solution for part a well since d we'll actually we type this out, make that a little nicer here.

01:39 Since d is equal to 10 centimeters, the image produced by the first lens will be 20 centimeters beyond the second lens.

02:09 So that's because if we go back here, q1 is 30 centimeters, but the distance is only 10.

02:16 Centimeters so 30 minus 10 is 20 so the the image produced by the first lens will be 20 centimeters beyond the second lens and that will allow us then to figure out where the object of the second lens is going to be and so that's actually our answer for part a and that leads us into or part b oh looks like i spelled beyond wrong there that's supposed to be an e okay and then part c actually is similar to part b, part c says what is the value p2 of the, which is the object position for the second lens.

03:03 So p2, since that object is to the right of the second lens, it's going to be virtual, therefore it's negative.

03:10 P2 is equal to negative 20 centimeters.

03:18 That can be boxed in as a solution for part c.

03:23 Part d then asks us to find the image that's produced by that second lens.

03:29 So if we go back to this page here, we're going to use the exact same equation as this.

03:36 But instead of using the one for the first lens, we're using two for the second lens.

03:41 So therefore, q2 is going to be equal to p2 times f2 divided by p2 minus f2...

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