how to draw the mapping and explain why draw like that? \4. The lens and mirror...
Consider the optical system shown in the figure. The lens and mirror are separated by d = 1.00 m and have focal lengths off, = +89.0 cm and f2 = -48.8 cm, respectively. An object is placed p = 1.00 m to the left of the lens as shown. Object Lens Mirror 1.00 m 1.00 m i (a) What is the distance (in cm) and location of the final image formed by light that has gone through the lens twice?...
A lens and a concave mirror are separated at 20.0 cm as shown in Figure . The lens and the mirror have focal lengths of 2.0 cm and 6.0 cm, respectively. Assume that an object is placed upright to the left of the lens. The image formed due to light passing through the lens twice is located at 6.0 cm to the right of the lens. (i) Determine the position of the object with respect to the lens. (ii) Describe...
The lens and mirror in the figure below are separated by 1.00 m and have focal lengths of +79.9 cm and -49.6 cm, respectively. If an object is placed 1.00 m to the left of the lens, where will the final image be located? q3 = cm State whether the image is upright or inverted, upright. inverted. Determine the overall magnification.
The object in the figure beside is mid-way between the lens and the mirror, which are separated by a distance d-25.0 cm. The magnitude of the mirror's radius of curvature is 20.0 cm, and the lens has a focal length of-16.7 cm. Lens Object Mirronr (A) We first study the image formed by the lens only (As the spherical (a) Describe the image formed by the lens (location, magnification, real or virtual and (b) Construct a ray diagram of this...
8. An object is placed 1 m in front of a convex lens with an 80 cm focal length. A convex mirror is then placed 1 m behind the lens. The radius of curvature of the mirror is 50 cm. 1. Where is the final image of the object (the image formed by light that has passed through the lens twice), with respect to the lens? 2. What is the total magnification of the final image? 3. Is the image...
Consider a spherical mirror and lens separated by 45 cm. The mirror is on the left with a focal length of 100 cm. The lens is on the right with a focal length of −20 cm. A 5 cm tall object is placed 20 cm to the left of the lens. a) If you only consider the rays that move to the right from the object, fully characterize the final image in the system. In other words provide final image...
In the situation shown the lens has a focal length -10 cm and the mirror has curvature radius +20 cm. The distance between adjacent dots on the diagram is 10 cm. An object is placed as shown 15 cm in front of the mirror. a) Use the formulas to locate the image formed by the mirror and the image subsequently formed by the lens. (Ignore the image formed directly by the lens.) Calculate the magnification of the final image relative...
A concave spherical mirror has a radius of curvature of 22 cm. Locate the images for object distances as given below. In each case, state whether the image is real or virtual and upright or inverted, and find the magnification. (If an answer does not exist, enter DNE. If an answer is infinity, enter INFINITY.) (a) p=11 cm image distance image orientation -Select- magnification 7 x cm DNE (b) p- 22 cm image distance image orientation magnification 22 cm real...
A converging lens with a focal length of 4.9 cm is located 20.9 cm to the left of a diverging lens having a focal length of -11.0 cm. If an object is located 9.9 cm to the left of the converging lens, locate and describe completely the final image formed by the diverging lens. a) Where is the image located as measured from the diverging lens? b) What is the magnification? c) Also determine, with respect to the original object...
11.87 A 1.00-cm-high object is placed 4.85 cm to the left of a converging lens of focal length 8.20 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 Take the image formed by the first lens to be the object for the second lens and apply the lens equation to each lens to locate the final image. cm 8.442...