anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Michele_Laino
  • Michele_Laino
here is the drawing of your problem: |dw:1434906888314:dw|
anonymous
  • anonymous
whoahh okie!! :)
Michele_Laino
  • Michele_Laino
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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More answers

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

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