An object is located 51 millimeters from a diverging lens.The object has a height of 13 millimeters and the image height is 3.5 millimeters.How far in front of the lens is the image located?
|(image distance)/(object distance)| = (image size)/(object size)|
3.5/13 = (image distance)/51 mm
Image distance = 51*3.5/13 = 13.7 mm
You do not need to worry about minus signs and whether or not the lens is diverging to use this formula.
13.73
thanks
an object 5 millimeters high is loccated 15 millimeters in front of a plane mirror. How far from the mirror is the image located?
To solve this problem, we need to use the lens equation:
1/f = 1/d₀ + 1/dᵢ
Where:
- f is the focal length of the lens,
- d₀ is the object distance (distance of the object from the lens),
- dᵢ is the image distance (distance of the image from the lens).
In this case, we have a diverging lens, which means the focal length is negative. Let's assume f = -x, where x is the magnitude of the focal length. We can rewrite the lens equation:
-1/x = 1/d₀ + 1/dᵢ
The object distance, d₀, is given as 51 millimeters. To find the image distance, dᵢ, we can use the magnification equation:
magnification (m) = -dᵢ / d₀ = image height / object height
Given that the image height is 3.5 millimeters and the object height is 13 millimeters, we can rearrange the equation to solve for dᵢ:
dᵢ = -m * d₀
Now we have all the necessary information to calculate the image distance, dᵢ:
dᵢ = -m * d₀ = - (3.5/13) * 51
Let's calculate it:
dᵢ = - (3.5/13) * 51
= - (0.269) * 51
≈ - 13.72 millimeters
Therefore, the image distance, dᵢ, is approximately -13.72 millimeters. Since the distance is negative, it means the image is formed on the same side as the object.