An object is located at x=0 . At x=2.50 cm is a converging lens with a focal length of 2.00 cm

An object is located at x=0 . At x=2.50 cm is a converging...

An object is located at $x=0 .$ At $x=2.50 \mathrm{~cm}$ is a converging lens with a focal length of $2.00 \mathrm{~cm}$, at $x=16.5 \mathrm{~cm}$ is an unknown lens, and at $x=19.8 \mathrm{~cm}$ is another converging lens with focal length $4.00 \mathrm{~cm}$. An upright image is formed at $x=39.8 \mathrm{~cm} .$ For each lens, the object distance exceeds the focal length. The magnification of the system is $6.84$. (a) Is the unknown lens diverging or converging? (b) What is the focal length of the unknown lens? (c) Draw a ray diagram to confirm your answer.

Key Concepts

Lens Sign Conventions

Sign conventions in geometrical optics are critical for correctly applying the thin lens equation. They determine the direction of distances (object distance versus image distance) and the nature of the images (real versus virtual). Consistent use of sign conventions ensures that the calculations of image positions, distances, and magnifications accurately reflect the physical setup of the lens system.

Ray Diagram

A ray diagram is a graphical tool used to trace the paths of light rays through optical systems. It helps in visualizing how lenses bend light to form images and is particularly useful for confirming the outcomes predicted by mathematical analysis. Ray diagrams illustrate object and image relationships, focal points, and the effects of multiple lenses, thereby providing an intuitive understanding of the optical behavior of the system.

Lens Magnification

Magnification for a lens explains how the size of the image relates to the size of the object and is given by m = -di/do for a single lens. In multi-lens systems, the overall magnification is the product of the individual magnifications from each lens. This concept is crucial for understanding how multiple lenses affect the final image size and orientation, as well as for ensuring that the computed magnification corresponds with the desired overall system performance.

Thin Lens Equation

The thin lens equation, 1/f = 1/do + 1/di, relates the focal length (f) of a lens to the object distance (do) and the image distance (di). It is a fundamental relation in geometrical optics used to determine where an image will form once an object is placed at a certain distance from a thin lens. This equation underpins the analysis of single lens systems and is extended in sequential lens setups by treating the image of one lens as the object for the next.

Sequential Lens System Analysis

In a multiple-lens system, each lens contributes to the final image formation. Sequential lens system analysis involves applying the thin lens equation step by step, where the image formed by one lens serves as the object for the next. Corrections for the spacing between lenses and proper handling of sign conventions are essential for accurately predicting where and how the final image will form.

Transcript

00:02 At thence 1, first find object distance or find image distance which is 1 over f1 minus 1 over p1 inverse equals 1 over 2 minus 1 over 2 .5 inverse equals 10 centimeters.

00:36 So we will have p2 equals s minus q1 equals s minus q1 equals s equals s 16 .5 minus 2 .5 minus 10 equals 4 centimeter now the second the lens 3 is located at 19 .8 centimeter so we will have q3 equals 39 .8 minus its location 19 .8 equals 20 centimeter.

01:47 Now object distance at lens 3 will be 1 over f3 minus 1 over q3 inverse equals 1 over 4 minus 1 over 20 inverse equals 5 centimeter.

02:28 Now we come back to thence 2 so we have q2 equals its location, 90 .8 centimeter minus 5 centimeter equals 14 .8 centimeter and so this is the distance.

03:21 So this minus 16 .5 centimeter will be the actual object distance 16 .5 centimeter.

03:49 So q2 comes out to be 14 .8 minus 16 .5 .5 equals negative 1 .7 centimeter.

04:02 Now we are trying to find the focal length of the second lens.

04:05 So f2 equals 1 over p2 plus 1 over q2 inverse of that equals 1 over 4 minor plus 1 over negative 1 .7 inverse equals negative 3 centimeter and the negative signifies it is a diverging length and focal length is three centimeter now if we want to draw the radar diagram we have to the first sense is converging lens you draw the optical axis if this is the focus and you start from here then just two rays will do it but let's just draw a third ray as well this way will keep going straight and here is where the first virtual in first real image form or first intermediate image is formed over here now this will serve as the object for the second lens which came out to be diverging lens and let's say this is the focus put one ray through the focus put one ray straight and the put other ray through the optical axis the ray that goes through the optical axis goes straight and here is the focus so this ray will keep going in this direction.

06:27 Since this is a diverging lens, you should not put a ray through the first focus...

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