Need help with #7 please **Lens Equation: Lab Activity**The relationship between the object distance \(d_o\), image distance \(d_i\), and the focal length \(f\) can be modeled by the thin lens equation:\[\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}\]A thin lens is a lens where the width of the lens is a lot smaller than the diameter of the lens.**Double convex lens:** Assume that the object distance \(d_o\) is positive if it’s to the left of the lens, the image distance \(d_i\) is positive if it’s to the right of the lens, the focal point is positive if it is on the other side of the lens as the object.**Double concave lens:** Assume that the object distance \(d_o\) is positive if it’s to the left of the lens, the image distance \(d_i\) is negative if it’s on the same side of the lens, and the focal point is negative if it’s on the same side as the object.**Focal length** is positive if working with a double convex lens and negative if working with a double concave lens.The lens magnification can be described using the formula \(m = -\frac{h_i}{h_o} = -\frac{d_i}{d_o}\) where \(h_o\) is the object’s height, and \(h_i\) is the image's height.**Simulation link:** [http://ophysics.com/l12.html](http://ophysics.com/l12.html)In this activity, you will use the thin lens equation to find the objects distance analytically and then verify it using the provided simulation. **Question 7:** Now imagine that you are working with a double concave lens. You can convert the convex lens to a double concave by moving the point labeled focus’ to the other side of the lens. Create the following scenario f = -2.0, object of height h₀ = 1 unit is placed at x = -8 and the object distance is d₀ = 8 units. What is the image distance, the magnification, type of image formed and the orientation of the image? Why is this image called virtual?**Question 8:** Verify the results from question 7 using the thin lens equation and the equation for magnification.

The object distance is negative for convex lenses, while the image distance is positive for real…

Question 7: Now imagine that you are working with a double concave lens. You can convert the convex lens to a double concave by moving the point labeled focus' to the other side of the lens. Create the following scenario f = -2.0, object of height họ = 1 unit is placed at x = -8 and the object distance is do = 8 units. What is the images distance, the magnification, type of image formed and the orientation of the image? Why is this image called virtual?

Question 7: Now imagine that you are working with a double concave lens. You can convert the convex lens to a double concave by moving the point labeled focus' to the other side of the lens. Create the following scenario f = -2.0, object of height họ = 1 unit is placed at x = -8 and the object distance is do = 8 units. What is the images distance, the magnification, type of image formed and the orientation of the image? Why is this image called virtual?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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