An object 1.5cm high 12.0cm from a diverting lens of focal length 3.0cm. Find the length and position of the image
To solve this problem, we can use the lens formula which states:
1/f = 1/v - 1/u
where:
f = focal length of the lens (3.0 cm)
v = distance of the image from the lens (unknown)
u = distance of the object from the lens (12.0 cm)
We can rearrange the formula to solve for v:
1/v = 1/f + 1/u
Now, let's substitute the given values:
1/v = 1/3.0 + 1/12.0
Simplifying the equation:
1/v = (4 + 1) / 12.0 = 5/12.0
Now, find the reciprocal of both sides:
v = 12.0/5 = 2.4 cm
So, the length of the image is 2.4 cm.
To find the position of the image, we can use the magnification formula which states:
m = -v/u
where:
m = magnification
v = distance of the image from the lens (2.4 cm)
u = distance of the object from the lens (12.0 cm)
Now, let's substitute the given values:
m = -2.4 / 12.0 = -0.2
Since the magnification is negative, it tells us that the image is inverted.
Therefore, the length of the image is 2.4 cm and its position is 0.2 times the distance of the object from the lens, in this case, -2.4 cm.
To find the length and position of the image formed by a diverging lens, we need to apply the lens equation:
1/f = 1/u - 1/v
where f is the focal length of the lens, u is the object distance from the lens, and v is the image distance from the lens.
Given:
Height of the object (h) = 1.5 cm
Distance of the object from the lens (u) = 12.0 cm
Focal length of the lens (f) = -3.0 cm (since this is a diverging lens)
Step 1: Find the image distance (v).
Substituting the given values into the lens equation:
1/-3 = 1/12 - 1/v
Simplifying the equation:
-1/3 = 1/12 - 1/v
-4v = v - 12
-5v = -12
v = 2.4 cm
The image distance (v) is 2.4 cm.
Step 2: Find the length of the image (h').
Using the magnification formula:
h'/h = -v/u
Substituting the given values:
h'/1.5 = -2.4/12
h' = (1.5)(-2.4/12)
h' = -0.3 cm
The length of the image (h') is -0.3 cm.
Note: The negative sign indicates that the image is virtual and upright.
To find the length and position of the image formed by a diverging lens, you can use the lens formula:
1/f = 1/v - 1/u
where:
f is the focal length of the lens,
v is the image distance, and
u is the object distance.
In this case, the object distance (u) is given as 12.0 cm and the focal length (f) is given as -3.0 cm (since it's a diverging lens with a negative focal length).
To find the image distance (v), rearrange the lens formula equation as:
1/v = 1/f + 1/u
Substituting the given values into the equation:
1/v = 1/(-3.0) + 1/12.0
Simplifying this equation, you get:
1/v = -1/3 + 1/12
Combining the fractions on the right side of the equation:
1/v = -4/12 + 1/12
1/v = -3/12
Now, invert both sides of the equation to get v:
v = 12/(-3)
v = -4 cm
The negative sign indicates that the image formed by the diverging lens is virtual and located on the same side as the object.
To find the length of the image, you can use the magnification formula:
magnification (m) = -v/u
Substituting the values into the formula:
m = -(-4) / 12
m = 4/12
m = 1/3
The magnification of the image is 1/3, which means the image is one-third the size of the object.
Since the original object is 1.5 cm high, the length of the image is:
Length of image = 1/3 * 1.5 cm
Length of image = 0.5 cm
Therefore, the length of the image formed by the diverging lens is 0.5 cm, and its position is 4 cm in front of the lens (on the same side as the object).