A 2.4 cm object is 15.0 cm to the left of a convex lens with a focal length of +5.0 cm. Part 1: Draw a ray diagram of the setup showing the location of the image. Part 2: Calculate the distance of the image from the lens. Part 3: What size is the image and is it upright or inverted?

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A 2.4 cm object is 15.0 cm to the left of a convex lens with a focal length of +5.0 cm. Part 1: Draw a ray diagram of the setup showing the location of the image. Part 2: Calculate the distance of the image from the lens. Part 3: What size is the image and is it upright or inverted?

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here is the drawing of your problem: |dw:1434906888314:dw|
whoahh okie!! :)
here we have to apply this formula: \[\Large \frac{1}{p} + \frac{1}{q} = \frac{1}{f}\] where p= 15 cm f= 5 cm please solve for q

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ok! 1/15 + 1/q = 1/5 1/q = 3 q=1/3 ?
we have: \[\Large q = \frac{{fp}}{{p - f}} = \frac{{5 \times 15}}{{15 - 5}} = ...cm\]
ohh oops so we get 75 / 10 = 7.5 cm ?
that's right!
now, from my drawing, we get that the image is inverted
okay! so part 2 is that it is 7.5 cm away from the lens? and part 3 is that it is inverted?
yes! nevertheless we have to determine the dimension of the image
yes! we have to calculate the size, too.. how do we do that?
we can use the similarity property of these triangles: |dw:1434907503742:dw|
ok!
so we can write this proportion: \[\Large 2.4:15 = x:7.5\]
or, applying the fundamental property of proportion: \[\Large x = \frac{{2.4 \times 7.5}}{{15}} = ...cm\]
we get 1.2 cm? that is the size of the image?
yes! correct!
yay! so we are finished with this? :O
yes! we have finished! :)
yay!! okay, onto the last one!!:) thank you1!
ok! :)

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