a converging lens has a focal length of 15cm. Find two positions where a real object can produce images four times its size.
Di = 4 Do, or
Di = -4 Do
will both work.
Di = image distance
Do = object distance
First case:
1/Di + 1/Do = 1/f = 1/15
1/(4Do) + 1/Do = 1/15
5/(4Do) = 1/15
Do = 75/4 = 18.75 cm
Second case:
-1/4Do + 1/Do = 1/15
3/(4Do) = 1/15
4 Do = 45
Do = 11.25 cm
To find the positions where a real object can produce images four times its size using a converging lens, we can use the lens formula:
1/f = 1/v - 1/u
Where:
- f is the focal length of the lens
- v is the distance of the image from the lens (positive for a real image on the same side as the object)
- u is the distance of the object from the lens (positive for an object on the same side as the observer)
Given that the focal length (f) of the lens is 15 cm, and we want to find the positions where the image is four times the size of the object, let's assume that the object's height (h) is h and the image's height is 4h.
Position 1: Object beyond the focal point (u > f)
In this case, the object will be placed further away from the lens than the focal length. We can set the object distance (u) as a variable and solve for the image distance (v) using the magnification equation:
h'/h = -v/u
Where:
- h' is the height of the image
- h is the height of the object
- v is the image distance
- u is the object distance
Given that h' = 4h and substituting the known values into the equation, we have:
4h/h = -v/u
4 = -v/u
Since we know that the focal length (f) is 15 cm, we can substitute it into the lens formula:
1/15 = 1/v - 1/u
By rearranging the equation, we can express v in terms of u:
1/v = 1/15 + 1/u
v = 1 / (1/15 + 1/u)
v = 15u / (u + 15)
Substituting v into the magnification equation, we have:
4 = - [15u / (u + 15)] / u
4 = - 15u / (u(u + 15))
By cross-multiplying and rearranging, we get:
4u(u + 15) = -15u
4u² + 60u + 15u = 0
4u² + 75u = 0
u(4u + 75) = 0
From this equation, we can solve for u. Considering u = 0 is not physically meaningful, we find:
4u + 75 = 0
4u = -75
u = -75 / 4
u ≈ -18.75 cm (approximately)
Therefore, for Position 1, the object should be placed approximately 18.75 cm in front of the lens (on the same side as the observer).
Position 2: Object between the focal point and the lens (0 < u < f)
In this case, the object will be placed between the lens and the focal point. We can follow a similar approach as in Position 1, but considering that the object's distance (u) is now positive.
Using the same magnification equation:
4 = - [15u / (u + 15)] / u
By cross-multiplying and rearranging, we get:
4u(u + 15) = 15u
4u² + 60u - 15u = 0
4u² + 45u = 0
u(4u + 45) = 0
Again, considering u = 0 is not physically meaningful, we find:
4u + 45 = 0
4u = -45
u = -45 / 4
u ≈ -11.25 cm (approximately)
Therefore, for Position 2, the object should be placed approximately 11.25 cm in front of the lens (on the same side as the observer).
Note: The negative values for u indicate that the object is on the same side as the observer, while positive values would indicate that the object is on the opposite side of the lens.