SOLUTION: Hi! I'm having some difficulty with the following problem: If F is the focal length of a convex lens and an object is placed at a distance x from the lens, then its image wil

Algebra ->  Equations -> SOLUTION: Hi! I'm having some difficulty with the following problem: If F is the focal length of a convex lens and an object is placed at a distance x from the lens, then its image wil      Log On


   

Question 196582: Hi!
I'm having some difficulty with the following problem:
If F is the focal length of a convex lens and an object is placed at a distance x from the lens, then its image will be at a distance y from the lens, where F, x, and y are related by the lens equation below. Suppose that a lens has a focal length of 2.1 cm, and that the image of an object is 4 cm closer to the lens than the object itself. How far from the lens is the object?
1/F = 1/x + 1/y
thanks!
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Since "the image of an object is 4 cm closer to the lens than the object itself", this means that y=x-4


1%2FF+=+1%2Fx+%2B+1%2Fy+ Start with the given equation.


1%2F2.1+=+1%2Fx+%2B+1%2F%28x-4%29 Plug in F=2.1 and y=x-4


Multiply EVERY term by the LCD 2.1x%28x-4%29 to clear out the fractions.


x%28x-4%29=2.1%28x-4%29%2B2.1x%281%29 Cancel out and simplify


x%28x-4%29=2.1%28x-4%29%2B2.1x Multiply


x%5E2-4x=2.1x-8.4%2B2.1x Distribute


10x%5E2-40x=21x-84%2B21x Multiply EVERY term by 10 to make every number a whole number.


10x%5E2-40x-21x%2B84-21x=0 Get all terms to the left side.


10x%5E2-82x%2B84=0 Combine like terms.


Notice we have a quadratic in the form of Ax%5E2%2BBx%2BC where A=10, B=-82, and C=84


Let's use the quadratic formula to solve for x


x+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


x+=+%28-%28-82%29+%2B-+sqrt%28+%28-82%29%5E2-4%2810%29%2884%29+%29%29%2F%282%2810%29%29 Plug in A=10, B=-82, and C=84


x+=+%2882+%2B-+sqrt%28+%28-82%29%5E2-4%2810%29%2884%29+%29%29%2F%282%2810%29%29 Negate -82 to get 82.


x+=+%2882+%2B-+sqrt%28+6724-4%2810%29%2884%29+%29%29%2F%282%2810%29%29 Square -82 to get 6724.


x+=+%2882+%2B-+sqrt%28+6724-3360+%29%29%2F%282%2810%29%29 Multiply 4%2810%29%2884%29 to get 3360


x+=+%2882+%2B-+sqrt%28+3364+%29%29%2F%282%2810%29%29 Subtract 3360 from 6724 to get 3364


x+=+%2882+%2B-+sqrt%28+3364+%29%29%2F%2820%29 Multiply 2 and 10 to get 20.


x+=+%2882+%2B-+58%29%2F%2820%29 Take the square root of 3364 to get 58.


x+=+%2882+%2B+58%29%2F%2820%29 or x+=+%2882+-+58%29%2F%2820%29 Break up the expression.


x+=+%28140%29%2F%2820%29 or x+=++%2824%29%2F%2820%29 Combine like terms.


x+=+7 or x+=+6%2F5 Simplify.


So the possible answers are x+=+7 or x+=+6%2F5

However, since 6%2F5-4=1.2-4=-2.8 (which doesn't make much sense), this means we'll ignore x+=+6%2F5


So the only answer is x+=+7


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Answer:

So the object is 7 cm from the lens.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
If F is the focal length of a convex lens and an object is placed at a distance x from the lens, then its image will be at a distance y from the lens, where F, x, and y are related by the lens equation below. Suppose that a lens has a focal length of 2.1 cm, and that the image of an object is 4 cm closer to the lens than the object itself. How far from the lens is the object?
1/F = 1/x + 1/y
y-4 = distance the image is from the lens
y = distance the object is from the lens
----
1/2.1 = 1/(y-4) + 1/y
Multiply thru by 2.1y(y-4) to get:
y(y-4) = 2.1y + 2.1(y-4)
y^2-4y = 2.1y + 2.1y - 8.4
y^2 - 8.2y + 8.4 = 0
10y^2 - 82y + 84 = 0
(10y-12)(y+7) = 0
y = 1.2 cm (distance the object is from the lens)
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Cheers,
Stan H.