a converging lens has a focal length of 15cm. Find two positions where a real object can produce images four times its size.
To find the two positions where a real object can produce images four times its size using a converging lens, we can use the magnification formula and the lens formula.
The magnification formula is given by:
magnification (m) = -image distance (di) / object distance (do)
Where:
- "di" is the distance of the image from the lens.
- "do" is the distance of the object from the lens.
In this case, the magnification is given as 4 (because the image is four times the size of the object).
So, we can write the magnification formula as:
4 = -di / do
Next, we can use the lens formula to establish a relationship between the object distance (do), image distance (di), and the focal length (f) of the converging lens:
1 / do + 1 / di = 1 / f
In this case, the focal length of the lens is given as 15 cm.
Now, let's find the two positions where the object can produce images four times its size.
Position 1:
Let's assume the object distance (do) is a positive value (since it's on the same side as the object) and the image distance (di) is a negative value (since it is on the opposite side of the lens).
Substituting the given values into the equations, we have:
4 = -di / do ----(1)
1 / do + 1 / di = 1 / f ----(2)
Substituting f = 15 cm into equation (2), we get:
1 / do + 1 / di = 1 / 15
Rearranging equation (2), we get:
di = 15do / (do - 15)
Now we can substitute this equation for "di" in equation (1) to solve for "do":
4 = -[15do / (do - 15)] / do
Simplifying the equation, we have:
4 = -15 / (do - 15)
Cross-multiplying, we get:
4(do - 15) = -15
Expanding and rearranging, we get:
4do - 60 = -15
4do = 45
do = 45 / 4
do ≈ 11.25 cm
So, the object distance (do) for position 1 is approximately 11.25 cm.
Now we can substitute this value of "do" back into equation (2) to find the image distance (di):
1 / (11.25) + 1 / di = 1 / 15
Simplifying the equation, we get:
1 / di = 1 / 15 - 1 / (11.25)
1 / di = (11.25 - 15) / (15 * 11.25)
di = (15 * 11.25) / (15 - 11.25)
di = 168.75 / 3.75
di = 45 cm
So, the image distance (di) for position 1 is 45 cm.
Position 2:
To find the second position, we can use the same steps as above but consider different values for "do".
Assuming "do" to be a different positive value, we can substitute that value into equations (1) and (2) to find the corresponding values of "di".
Following the same calculations, we can find the second position where the object can produce an image four times its size.
By solving these equations, we can find the appropriate values for "do" and "di" for position 2.
Therefore, the two positions where a real object can produce images four times its size using a converging lens are at approximately do = 11.25 cm and di = 45 cm.