When an object is placed 55.0cm from a certain converging lens, it forms a real image. When the object is moved to 45.0cm from the lens, the image moves 6.00 cm farther from the lens.
Find the focal length of this lens?
Please someone help me out I tried so many times i keep getting wrong answer.
PHYSICS PLEASE HELP ASAP - bobpursley, Wednesday, March 27, 2013 at 4:45pm
1/f=1/di+1/do
f=dido/(di+do)
but f before the move is f after the move, so
(di+6)(55+45)/(di+55+51)=di55/(di+55)
solve for di. Looks like fun.
then, after you get di, go for f from the lens equation
PHYSICS PLEASE HELP ASAP - Laila, Wednesday, March 27, 2013 at 6:48pm
I found this equation 5380di^2+600di+33000=0 and when i solve i got an imaginary number I had a hard time solving it so could you please help me out I actually redid it and got 3.15 for di i don't know if that's right or not an how to go from there cuz i keep gettind it wrong
To find the focal length of the converging lens, we can use the lens formula:
1/f = 1/di + 1/do
Where:
- f is the focal length of the lens
- di is the image distance
- do is the object distance
We are given that when the object is placed 55.0 cm from the lens, it forms a real image. Let's assign this as the object distance (do) for the first case:
do = 55.0 cm
We are also given that when the object is moved to 45.0 cm from the lens, the image moves 6.00 cm farther from the lens. This means that the image distance (di) changes by 6.00 cm. So, we can write the second case as:
do = 45.0 cm
di = di + 6.00 cm
Now, let's substitute these values into the lens formula:
1/f = 1/(di + 6.00) + 1/45.0
Next, we need to solve this equation for di. To do this, we can multiply both sides of the equation by (di + 6.00)(45.0) to simplify:
45.0(di + 6.00) + (di + 6.00)(di) = (di)(di + 6.00)(45.0)
Expanding and rearranging the equation:
45.0di + 270.0 + di^2 + 6.00di = 45.0di^2 + 270.0di
Rearranging again:
45.0di^2 + (270.0 - 6.00)di - (45.0di + 270.0) = 0
Combining like terms:
45.0di^2 + 264.0di - 45.0di - 270.0 = 0
Simplifying further:
45.0di^2 + 219.0di - 270.0 = 0
Now, we have a quadratic equation in terms of di. We can solve this equation using the quadratic formula:
di = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = 45.0, b = 219.0, and c = -270.0.
di = (-219.0 ± √(219.0^2 - 4 * 45.0 * -270.0)) / 2 * 45.0
Simplifying the equation inside the square root:
di = (-219.0 ± √(47961.0 + 48600.0)) / 90.0
di = (-219.0 ± √(96561.0)) / 90.0
Calculating the square root:
di = (-219.0 ± 310.6) / 90.0
Now, we have two possible values for di:
di1 = (-219.0 + 310.6) / 90.0 = 3.0192 cm (approximately)
di2 = (-219.0 - 310.6) / 90.0 = -5.4292 cm (approximately)
Since a real image is formed, we can discard the negative value for di. Therefore, the image distance is approximately 3.0192 cm.
Now, we can substitute this value of di back into the lens formula to find the focal length:
1/f = 1/(3.0192 + 6.00) + 1/45.0
Simplifying:
1/f = 1/9.0192 + 1/45.0
Adding the fractions:
1/f = (45.0 + 9.0192) / (9.0192 * 45.0)
1/f = 54.0192 / 406.864
Inverting both sides of the equation:
f = 406.864 / 54.0192
Evaluating this expression:
f ≈ 7.53 cm (approximately)
Therefore, the focal length of the converging lens is approximately 7.53 cm.