An apple is placed 17.0 cm in front of a diverging lens with a focal length of magnitude 23.2 cm. What is the image distance i of the image of the apple through this lens?
What is the magnification of the image of the apple?
Wwhich of the following statements are true and which are false?
The image of the apple is:
behind the lens?
virtual?
inverted?
unchanged in size?
To find the image distance i of the image of the apple, we can use the lens equation:
1/f = 1/d_o + 1/d_i
where f is the focal length, d_o is the object distance, and d_i is the image distance.
Given:
f = 23.2 cm (focal length)
d_o = -17.0 cm (negative sign indicates object is in front of the lens)
Plugging in these values, we can solve for d_i:
1/23.2 = 1/-17.0 + 1/d_i
0.0431034483 = -0.0588235294 + 1/d_i
1/d_i = 0.0431034483 + 0.0588235294
1/d_i = 0.1019269777
d_i = 9.814 cm (rounded to three decimal places)
Therefore, the image distance i of the image of the apple is approximately 9.814 cm.
To find the magnification of the image, we can use the magnification formula:
magnification (m) = -d_i / d_o
Given:
d_i = 9.814 cm (image distance)
d_o = -17.0 cm (object distance)
Plugging in these values, we can calculate the magnification:
m = -9.814 / (-17.0)
m = 0.577 (rounded to three decimal places)
Therefore, the magnification of the image of the apple is approximately 0.577.
Now, let's analyze the statements:
1. The image of the apple is behind the lens.
False. Since the object distance (d_o) is negative (-17.0 cm), the image is formed on the same side as the object (in front of the lens), making it a virtual image.
2. The image of the apple is virtual.
True. As mentioned above, the image formed is virtual because it is formed in front of the lens, making it impossible to project onto a screen.
3. The image of the apple is inverted.
True. The negative magnification value indicates an inverted image.
4. The image of the apple is unchanged in size.
False. The magnification value is not equal to 1, so the image is not the same size as the object.
To find the image distance, we can use the lens formula:
1/f = 1/di - 1/do
where f is the focal length of the lens, di is the image distance, and do is the object distance. In this case, the object distance is the distance of the apple from the lens, which is 17.0 cm. The focal length of the lens is given as 23.2 cm.
Plugging in the values:
1/23.2 = 1/di - 1/17.0
To solve for di, we can rearrange the equation:
1/di = 1/23.2 + 1/17.0
Now we can calculate di:
1/di = (17 + 23.2)/(17*23.2)
di = 17*23.2 / (17 + 23.2)
Using a calculator, we find di ≈ 10.58 cm.
Now let's calculate the magnification:
Magnification (m) is given by the formula:
m = -di / do
Plugging in the values:
m = -10.58 / 17
m ≈ -0.62
The negative sign indicates that the image is inverted.
Now let's analyze the statements:
1. The image of the apple is behind the lens: True. Since the image distance (di) is positive, it means the image is formed on the opposite side of the lens from the object, which is behind the lens.
2. The image of the apple is virtual: True. A diverging lens always produces a virtual image when the object is placed in front of it. In this case, the object distance (do) is positive, so the image is virtual.
3. The image of the apple is inverted: True. The negative magnification value indicates an inverted image.
4. The image of the apple is unchanged in size: False. The magnification value of -0.62 indicates that the image is reduced in size.
In summary, the correct statements are:
1. The image of the apple is behind the lens.
2. The image of the apple is virtual.
3. The image of the apple is inverted.
4. The image of the apple is not unchanged in size.