An object is placed 10 m before a convex lens with focal length 5.7 m . Another

Question

An object is placed 10 m before a convex lens with focal length 5.7 m . Another convex lens is placed 16.2 m behind the first lens with a focal length 11 m (see the figure below). [Note: Make a ray diagram sketch in order to check your numerical answer.]

a) At what distance is the first image from the first lens?

correct answer: 13.2558 m

b) What is the magnification of the first image?

correct answer: -1.32558

c) At what distance is the second image from the second lens?

correct answer: 4.02021 m

d) What is the magnification of the final image, when compared to the initial object?

correct answer: -1.81005

Thank You!

Explanation / Answer

What you want to do is solve the image distance for the first lens, and then use this image as the object for the second lens. *my answers aren't exactly the ones given because I didn't use as many sig. figs. 1) di=fdo/(do-f)=5.7m*10m/(10m-5.7m)=13.3m --the magnification=-di/do=-(13.3m/10m)=-1.33x 2)this image is formed 16.2m-13.3m=2.9m to the left of lens 2. This is your new do. Now just solve like you did for the first lens, using the focal point for lens 2 and this new do: di=fdo/(do-f)=11m*2.9m/(2.9m-11m)=-3.94m m=-di/do=-(-3.94m/2.9m)=1.36x --the total magnification is the product of m1 and m2: -1.33*1.36=-1.81x

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