A 2.11 cm high insect is 1.31 m from a 125 mm focal-length lens. Where is the image?
Distance of the image is ?
How high is it?
What type is it?
To find the position of the image, we can use the thin lens formula:
1/f = 1/u + 1/v
Where:
f = focal length of the lens
u = object distance
v = image distance
Given:
f = 125 mm = 12.5 cm (since 1 cm = 10 mm)
u = 131 cm (since 1 m = 100 cm)
To find the image distance (v), we rearrange the formula:
1/v = 1/f - 1/u
1/v = 1/12.5 - 1/131
Now we can calculate v using this equation.
To find the height of the image, we can use the magnification formula:
m = -v/u
Given:
u = 2.11 cm
To find the magnification (m), we can substitute the values into the formula and calculate.
Finally, to determine the type (magnification positive or negative) of the image, we look at the sign of the magnification (m). If it is positive, the image is upright (virtual). If it is negative, the image is inverted (real).
Now, let's calculate the values to find the image distance, height, and type.
1/v = 1/12.5 - 1/131
v = 11.96 cm (rounded to 2 decimal places)
m = -v/u
m = -(11.96/2.11)
m = -5.67 (rounded to 2 decimal places)
The image is located approximately 11.96 cm from the lens. The image height can be found by multiplying the object height by the magnification, which in this case is -5.67. Since the object height is not mentioned in the question, we cannot determine the exact height of the image.
Based on the negative magnification value (-5.67), the image formed by the lens is inverted (real).