A converging lens with a focal length of 9.00 cm forms an image of a 4.00-mm-tall real object that is to the left of the lens. The image is 1.30 cm tall and erect. Where are the object and image located? Is the image real or virtual?
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A converging lens with a focal length of 9.00 cm forms an image of a 4.00-mm-tall real object that is to the left of the lens. The image is 1.30 cm tall and erect. Where are the object and image located? Is the image real or virtual?
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- A 1.0-cmcm-tall object is 9.0 cmcm in front of a converging lens that has a 40 cmcm focal length. Calculate the image position. Express your answer with the appropriate units. Enter positive value if the image is on the other side from the lens and negative value if the image is on the same side as the object. Calculate the image height. Express your answer with the appropriate unitsA 2.0 cm tall object is 40 cm in front of a converging lens that has a 20 cm focal length. Use ray tracing to find the position and height of the image, using a ruler or paper with a grid. Calculate the image position and height and compare with your answer to (a). Is the image real or virtual?A 1.00-cm-high object is placed 4.30 cm to the left of a converging lens of focal length 8.45 cm. A diverging lens of focal length -16.00 cm is 6.00 cm to the right of the converging lens. Find the position and height of the final image. position height cm ---Select--- cm
- 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.…Is the final image real or virtual? Is the final image upright or inverted?For safety reasons, you install a rear-window lens with a -0.299 m focal length in your van. Before putting the van in reverse, you look through the lens and see the image of a person who appears to be 0.339 m tall and 0.243 m behind the van. Determine the following. (a) actual distance of the person behind the van m (b) height of the person
- A 7.75 mm high chocolate chip is placed on the axis of, and 13.9 cm from, a lens with a focal length of 6.91 cm. If it can be determined, is the chocolate chip's image real or virtual? virtual real cannot be determined How high is the image? Express the answer as a positive quantity. image height: mmA farsighted person can read printing as close as 25.0 cm when she wears contacts that have a focal length of 38.0 cm. One day, however, she forgets her contacts and uses a magnifying glass, as in the drawing. It has a maximum angular magnification of 7.42 for a young person with a normal near point of 25.0 cm. What is the maximum angular magnification that the magnifying glass can provide for her? Virtual Magnifying glass Object LE ho dj. imageA 1.0-cm-tall object is 70 cm in front of a converging lens that has a 30 cm focal length. Calculate the image position. Calculate the image height. Type a positive value if the image is upright and a negative value if it is inverted.