A convex lens has a focal length of 8 cm. An object that is 4 cm tall is placed at unknown distance from the lens and produces a virtual image that is 6 cm away from the lens. What is the distance of the object from the lens? What is the magnification? Is the image upright or inverted?
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- A convex lens has a focal length of 8 cm. An object that is 4 cm tall is placed at unknown distance from the lens and produces a virtual image that is 6 cm away from the lens.
- What is the distance of the object from the lens?
- What is the magnification?
- Is the image upright or inverted?
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- A lightbulb is 4.0 m from a wall. Part A: What is the focal length of a lens what will form an image on the wall that is twice the size of the lightbulb? Part B: What is the position (measured from the bulb) of the lens? Thank you in advance!Two converging lenses, each of focal length 14.9 cm, are placed 39.6 cm apart, and an object is placed 30.0 cm in front of the first lens. Where is the final image formed? The image is located cm ---Location--- in front of the first lens. in front of the second lens. behind the second lens. What is the magnification of the system? M = ✕Two lenses are separated by 50 cm. Lens 1 is convex and has a radius of curvature of magnitude 30 cm. Lens 2 is concave and has a radius of curvature of magnitude 40 cm and is located to the right of Lens 1. An object is located 20 cm to left of lens 1. Find the location of the final image of the object. What is the magnification of the final image?
- 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.…A diverging lens has a focal distance of f=0.2 cm. If an object is placed 0.1cm away, What is the image distance? Is the image real or virtual? What is the magnification?
- A thin lens has a focal point located 16.8 cm from the lens surface. Determine the object location which would produce an upright image which is exactly five times larger than the object.A 3 cm tall object is placed 2 cm to the left of a converging lens that has a focal length with a magnitude of 4 cm. A diverging lens with a focal length of magnitude 8 cm is placed 10 cm to the right of the first lens. What is the magnification of the final image produced by these two lenses? Make sure you say whether it is bigger/smaller and upright/inverted.A converging lens of focal length f=2 cm is used to focus the image of an object onto a screen. The object and screen are separated by 15 cm. The lens is placed between the object and the screen, at a distance from the screen such that the image of the object is focused into the screen. At what distance from the screen must the lens be placed in order to have an image magnification < 1? Provide your answer to two significant figures. distance = cm.
- A 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 student's uncorrected near and far points are at 9.0 cm and infinity, respectively. She looks through a meniscus-shaped lens as shown in the figure. The absolute value of the radius of curvature is 2.4 cm on the side of the lens near her eye, and 2.6 cm on the side away from her eye. The index of refraction of the lens is 1.4. What is the closest an object can get to the lens and still be clearly visible to the student? [You may assume that the lens is very close to her eye.] eye 3.5 cm 78.0 cm 10.2 cm 8.1 cm 2.3 cmA small object is placed to the left of a convex lens and on its optical axis. The object is 30 cm from the lens, which has a focal length of 15 cm. If the object is moved to a position 22 cm away, what will the image position be? (A positive value for p, or position, means the image is on the right side of the lens.)