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FC X C https:// MasteringPhysics: Homework 11-Google Chrome https:// session. masteringphysics.com /myct/itemView ?assignment ProblemID-59361554&offset;: next Google For eac PHY122 Physics for the Life Sciences ll Spring 2016 I glose Signed in as Mark Szkolnicki Help Homework 11 Enhanced EOC: Problem 18.73 Resources T previous l 17 of 18 l next Enhanced EOC: Problem 18.73 Correct About 26 A 3.0 cm -tall candle flame is 2.0 m from a wall You happen to have a lens with a focal length of 42 A 1.0 cm Part B www.che You may want to review (DA pages 588-592) For each location, what is the height of the image? For each For help with math skills, you may want to review: the orienta Express your answer using two significant figures. Enter your answers numerically separated by a comma. Solving Quadratic Equations A 1.0 O For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor www.che Solution o Using a ens to create an image. cm. For each you put th Subm My Answers Give Up A 4.0c Part C www.che For each location, what is the orientation of the image? A 4.0cm length of 4 O both inverted O one inverted, the other upright physic O both upright https Ilan Dec 8, 20 Subm My Answers Give Up 4:09 PM m Cortana. Ask me anything 4/19/2016Explanation / Answer
As the focal length is positive we do KNOW that it represents a convex lens ( a concave lens has a negative focal length )
1/d0 + 1/di = 1/f
and d0 + di = 200 cm
This is a simultaneous equation.
di = 200 - do from equation 2
now substitute into 1.
1/d0 + 1/( 200-d0) = 1/42
multiply both sides by do(200- d0)
200 - d0 + d0 = (200-d0)d0 / 42
simplify and multiply bs by 42
200 * 42 = 200 d0 - d0^2
turn into a standard quadratic.
d0^2 - 200 d0 + 8400 = 0
On solving, do = 140, 60
From this you find that d0 must be 140 or 60 from the lens and the wall must be 60 or 140 cm from that lens,
You will notice that due to symmetry both pairs of distances have the same values.
ie. if you swap which one is the object and image you must always get a valid solution.
And this is true for any optical system.
Now For do = 160 cm, di = 40 cm, ho = 3.0 cm
-di/do = hi/ho
hi = -(40/160)*3 = -0.75 (inverted)
Also for do = 40 cm, di = 160 cm, ho = 3.0 cm
-di/do = hi/ho
hi = -(160/40)*3 = -12.0 cm (inverted)
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