An object with height y =15.0 cm is placed on the optical axis in front of a converging lens with an object distance p= 45.0 cm . The converging lens produces a virtual image that is magnified by m =+5/3=1.67 . The image is on the incident light side of the lens. (Pay attention to +/- signs): What is the location of the image, i ? b. What is the focal length of the lens, f ? What is the height, y' of the image (including the correct sign)? а. с.

An object is placed 45.0cm to the left of lens A. Lens B is placed 40.0cm to the right of lens A. Lens A is converging and has a focal length of 20.0cm . Lens B is converging and has a focal length of 10.0cm . a. What is the final image position relative to lens B? (Include the + or – sign.) b. What is the overall magnification? (Include the + or – sign.) c. Is the final image real or virtual? d. Is the final image upright or inverted?

Create a ray diagram for eyeglasses that contain a diverging lens. Assume you are looking at a 2 cm tall object that is 4 cm from the lens. The focal length is 3 cm. What type of vision are they used to correct? Give both names. How do they do this? Explain.

It is your first day at work as a summer intern at an optics company. Your supervisor hands you a diverging lens and asks you to measure its focal length. You know that with a converging lens, you can measure the focal length by placing an object a distance ss to the left of the lens, far enough from the lens for the image to be real, and viewing the image on a screen that is to the right of the lens. By adjusting the position of the screen until the image is in sharp focus, you can determine the image distance s′s′ and then use the equation 1s+1s′=1f1s+1s′=1f, to calculate the focal length ff of the lens. But this procedure won'’t work with a diverging lens−−by itself, a diverging lens produces only virtual images, which can'’t be projected onto a screen. Therefore, to determine the focal length of a diverging lens, you do the following: First you take a converging lens and measure that, for an object 20.0 cmcm to the left of the lens, the image is 29.7 cmcm to the right of the lens.…

Thin lenses. Object O stands on the central axis of a thin symmetric lens. For this situation, each problem in the table (below) gives object distance p (centimeters), the type of lens (C stands for converging and D for diverging), and then the distance (centimeters, without proper sign) between a focal point and the lens. Find (a) the image distance i and (b) the lateral magnification m of the object, including signs. Also, determine whether the image is (c) real or virtual, (d) inverted from object O or noninverted, and (e) on the same side of the lens as object O or on the opposite side. (a) (b) (c) (d) (e) Lens i m R/V I/NI Side +14 C, 24 (a) Number Units (b) Number i Units (c) (d) (e) > >

A converging lens has a focal length of 20.0 cm. Construct accurate ray diagrams for object distances of (i) 60.0 cm and (ii) 6.67 cm. (a) From your ray diagrams, determine the location of each image. Image (i) distance = Image (ii) distance =