SPH3UW Unit 7.8 Multiple Lens Page 1 of 5

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1 SPH3UW Unit 7.8 Multiple Lens Page of 5 Notes Physics Tool box Thin Lens is an optical system with two refracting surfaces. The most simplest thin lens contain two spherical surfaces that are close enough together that we can neglect the distance between them. Converging Lens a lens that has the rays converge to a single point, such as a double convex lens. Diverging Lens a lens that has parallel rays diverge, thus giving the lens a negative value, such as a double concave lens. Object-Image Relation - so si f hi si Lateral Magnification - m h s s o o If two thin lenses are used in combination to form an image, the system can be treated in the following way: If you have two thin lenses that are used to form an image, you can treat the system in the following manner: First, the image formed by the first lens is located as if the second lens were not present at all. Second, a ray diagram is drawn for the second lens, using the image formed by the first lens as the object for the second lens. The final image is the image location of the system. Double Lens Technique If the image formed by the first lens lies behind the second lens, then that image is treated as a virtual object for the second lens and you apply the Burns-Schlueter Theroem (BST) for ray tracing, and when applying the lens equation, the object distance, so, is treated as negative. Since the magnification due to the second lens is performed on the magnified image due to the first lens, the overall magnification of the image due to the combination of lenses is the product of the individual magnifications.

2 SPH3UW Unit 7.8 Multiple Lens Page of 5 Burns-Schlueter Theorem: In a multiple lens system, when the image of the first lens lies behind the second lens, you treat the second lens as a lens of its negative focal length. That is, a converging lens becomes a diverging lens, or a diverging lens can be treated as a converging lens. The ray drawing techniques are applied as usual, and the final image location will be located. Image from First lens. This will be used as a virtual object for the second lens. The two rays intersect here, so this is the final image location Thru the centre ray. The parallel ray now diverges away from the second lens, via BST.. The object s (virtual object) ray thru the centre will continue along a straight line.. Extend this ray backwards for possible image location 3. The ray parallel to principle axis 4. The ray now is treated as if it was passing thru a diverging lens of focal length f 5. Trace ray 4 backwards to see if and where it intersects ray.

3 SPH3UW Unit 7.8 Multiple Lens Page 3 of 5 Example The figure below shows the two lenses and a ray-tracing diagram. From the ray-tracing diagram, we find that the final image is approximately 50 cm from the second lens and the height of the final image is 4.5 cm. The ray-tracing shows that the lens combination will produce a real, inverted, and larger image behind the second lens. To location the final image using algebra: S = 5 cm is the object distance of the first lens. Its image,, from the left lens is a virtual image, is found from the thin-lens equation: f s 40cm 5cm 5 0cm 4cm The negative sign indicates image on same side as original object.

4 SPH3UW Unit 7.8 Multiple Lens Page 4 of 5 The magnification due to the first lens is: m 4cm s 5cm.6 The positive sign indicates image is oriented in the same direction as the object (up). The image,, of the first lens is now the object for the second lens. The object distance from the second is s = 4 cm + 0 cm = 34 cm. A second application of the thin-lens equation yields: f s 0cm 34cm 4 680cm 48.6cm The positive sign indicates image on opposite side as original object and is about 49 cm from the second lens (right most). The magnification due to the second lens is: m 48.6cm s 34cm.49 The positive sign indicates image is oriented in the opposite direction as the virtual object (the first image). The combined magnification is M = m m = (.6)(.49) =.86. The height of the final image is (.86)(.0 cm) = 4.6cm and it is inverted from the original object.

5 SPH3UW Unit 7.8 Multiple Lens Page 5 of 5 Extra Notes and Comments