the correct lens formula is 1/f = 1/v+1/u so why have you rearranged as 1/v - 1/u = 1/f? I do not understand why 25cm is negative, please draw this on a lens diagram to help aid my understanding of when to use sign conventions
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the correct lens formula is 1/f = 1/v+1/u so why have you rearranged as 1/v - 1/u = 1/f?
I do not understand why 25cm is negative, please draw this on a lens diagram to help aid my understanding of when to use sign conventions
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Solved in 3 steps with 1 images
- The diagram shows 2 out of the 3 possible principal rays traversing a thin lens of unknown type. The focal points are labeled F1 and F2, the object is at 0, and the image is at I. Which of the following statements are true? Select one or more: O a. This is a divergent lens. Ob. This is a convergent lens. Oc. The magnification is about-2 Od. The magnification is about 2. De. The image is virtual. Of The image is realxviA small insect is placed 5.40 cmcm from a 6.00-cmcm-focal-length lens. a) Calculate the position of the image. Follow the sign conventions. Express your answer using two significant figures and include the appropriate units. b) Calculate the angular magnification. Express your answer using two significant figures.
- Could you please explain me why is not in front but behind? How is it possible that a divergent lens can form a real immage, shouldn t be impossible?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.HW Q7
- 30 cm f₁=10 cm 25 cm f₂=30 cm A converging lens (f = 10 cm) and a second converging lens (f = 30 cm) are placed 25 cm apart, and an object is placed 30 cm in front of the first converging lens. Object's height is 2 cm. Where is the location of the final image? 7.5 cm away from lens 2 and on the right side of lens 2. O 15 cm away from lens 2 and on the right side of lens 2. 30 cm away from lens 2 and on the right side of lens 2. 30 cm away from lens 2 and on the left side of lens 2. 7.5 cm away from lens 2 and on the left side of lens 2. 15 cm away from lens 2 and on the left side of lens 2.Please help me with the following question, explain and make sure it's correct, thank you sm!!An object is held 76.7 cm away from a lens. The lens has a focal length of 78.3 cm. A.) Determine the image distance. -3753 cm B.) Determine the magnification. 48.93 C.) If the object is 6.3 cm long, determine the image height. 308.25 cm D.) Which of the following are true about the image? Choose all that are true. O The image is magnified. u The image is inverted. The image is virtual. - The image is diminished. The image is real. - The image is upright. E.) Which of the following are true about the lens? Choose all that are true. The lens is concave. The lens is diverging. M The lens is converging. M The lens is convex.