An object with a height of 7.2 mm is located 37.9 cm to the left of lens [tex]f_1[/tex] with a focal length of -30 cm. At a distance of 83 cm to the right of lens [tex]f_1[/tex] is lens [tex]f_2[/tex] with a focal length of 36 cm. Determine the magnification of the final image.

First, we calculate the image distance for the first lens. Then, taking that image as the object for the second lens, we calculate the image distance for the second lens. The total magnification is the product of the magnifications of the two lenses, which is -1.28.

Explanation:

The magnification of an image through a lens is given by the negative ratio of the image distance to the object distance. First, we find the image distance for the first lens using the lens formula 1/f = 1/u + 1/v. Here, f is the focal length, u is the object distance, and v is the image distance. F in this problem is -30 cm, and u is -37.9 cm (because object lies on left of lens). Solving this for v, we get an image distance of -64.4cm.

Next the new object is 83cm - (-64.4cm) = 147.4cm, to the right of lens f₂ with a focal length of 36cm. Repeat the lens formula to find the new v, we get v = 42cm. Thus, the total magnification is the product of the magnification of the two lenses, which is m = -v₁/u₁ * -v₂/u₂ = 64.4/37.9 * -42/147.4 = -1.28

Learn more about Lens magnification here:

https://brainly.com/question/29355532

#SPJ11