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.
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
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
To find the focal length of the lens, we can use the lens formula:
1/f = 1/d₀ - 1/dᵢ
Where:
f is the focal length of the lens,
d₀ is the object distance from the lens, and
dᵢ is the image distance from the lens.
We have two sets of data:
When the object is placed at a distance of 55.0 cm from the lens, a real image is formed.
So, d₀ = 55.0 cm.
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.
So, d₀ = 45.0 cm and d₁ = dᵢ + 6.00 cm.
Let's substitute these values into the lens formula and solve for f:
1/f = 1/d₀ - 1/dᵢ [Equation 1]
1/f = 1/55 - 1/(dᵢ + 6) [Equation 2]
Now, we can solve this equation to find the value of f.
Subtracting Equation 2 from Equation 1:
0 = 1/55 - 1/(dᵢ + 6) - 1/55 + 1/dᵢ
0 = -1/55 - 1/(dᵢ + 6) + 1/dᵢ
Cross multiplying:
dᵢ = -6
Now that we have the value of dᵢ, we can substitute it into Equation 1 to find the focal length:
1/f = 1/55 - 1/(dᵢ + 6)
1/f = 1/55 - 1/(-6 + 6)
1/f = 1/55
Thus, the focal length of the lens is 55.0 cm.