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 lens, you 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
You are given that when the object is at a distance of 55.0 cm from the lens, it forms a real image. Let's call this di1. When the object is moved to a distance of 45.0 cm from the lens, the image moves 6.00 cm farther from the lens. This means the new image distance is di2 = di1 + 6.
Now, let's substitute these values into the lens formula:
1/f = 1/di2 + 1/do
Substitute di2 = di1 + 6:
1/f = 1/(di1 + 6) + 1/do
Rearrange the equation to solve for do:
1/do = 1/f - 1/(di1 + 6)
Now, you are given that when the object is at a distance of 55.0 cm from the lens, it forms a real image. This means the image distance (di1) is negative since it is on the opposite side of the lens. Substituting this value:
1/do = 1/f - 1/(-55 + 6)
Simplify:
1/do = 1/f + 1/49
Now, let's consider the second situation when the object is at a distance of 45.0 cm from the lens, and the image distance is di2 = di1 + 6. Substitute these values into the lens formula:
1/f = 1/di2 + 1/do
1/f = 1/(di1 + 6) + 1/45
Again, substitute the value of di1:
1/f = 1/(-55 + 6 + 6) + 1/45
Simplify:
1/f = 1/(-43) + 1/45
Now you have two equations:
1/do = 1/f + 1/49
1/f = 1/(-43) + 1/45
Multiply the second equation by (-43) * 45 to get rid of the fractions:
45/f = 45/45 - 43/45
Simplify:
45/f = 2/45 ⇒ 1/f = 2/45 * 1/45
Multiply the two equations:
(1/do) * (1/f) = (1/f + 1/49) * (1/f)
This simplifies to:
1/(do * f) = (1/f + 1/49) * (1/f)
Multiply both sides by (do * f):
1 = (1/f + 1/49) * (1/f) * (do * f)
1 = (1/f + 1/49) * (1/f) * do
Now, substitute the value of 1/f from the previous equation:
1 = (2/45 * 1/45 + 1/49) * (2/45 * 1/45) * do
Simplify:
1 = (2/2025 + 1/49) * (2/2025) * do
Find the common denominator:
1 = (98/2025 + 1/49) * (2/2025) * do
1 = (98/2025 + 45/2025) * (2/2025) * do
Simplify and combine the fractions:
1 = 143/2025 * (2/2025) * do
1 = 286/414_062_500 * do
Multiply both sides by 414_062_500 to isolate do:
414_062_500 = 286 * do
Divide both sides by 286 to find do:
do = 414_062_500 / 286
do ≈ 1_448_675.874
Now, you have the value of do. Substitute this value into either of the lens formulas to solve for f.
Let's use the first equation:
1/f = 1/(di1 + 6) + 1/do
Substitute the values:
1/f = 1/(-55 + 6) + 1/1_448_675.874
Simplify:
1/f = 6.114066025 * 10^-5
Now, solve for f:
f = 1 / (6.114066025 * 10^-5)
f ≈ 16343.3367
Therefore, the focal length of the lens is approximately 16,343.3367 cm.