A lamp 10 cm high is placed 60cm in front of a diverging lens of focal length 20 cm. By means of a scale ray digram, locate the image and determine the height.
I'm not too sure how i would find the distance of the image, as it is a diverging lens
The image would be virtual with a diverging lens. Use -20 cm for the focal length
-1/20 = 1/di + 1/do = 1/di + 1/60
1/di = -0.0500 - 0.0167 = -0.0667
di = -15 cm
The image height is 15/60 * 10 = 2.5 cm
We can't draw the ray diagram for you. it isn't hard to do. Parallel rays from the top and bottom of the lamp, after intersection, will intersect at the focal point in front of the lens.
I used, 1/f = 1/di +1/do
1/di = (1/20) - (-1/60)
= 15cm
but shouldn't the answer be -15 and not 15?
To locate the image produced by a diverging lens, you can use the scale ray diagram and follow these steps:
Step 1: Draw a horizontal line to represent the principal axis of the lens.
Step 2: Place the diverging lens on the principal axis. Since the lens is diverging, the left side of the lens will be thicker.
Step 3: Draw an optical center (O) inside the lens, which is usually located near the center of the lens.
Step 4: Place the object (lamp) vertically above the principal axis and mark its height (10 cm) with a vertical arrow pointing downward from the top.
Step 5: Draw three primary rays from the top of the object through the lens:
a) The first ray is parallel to the principal axis and refracts away from the principal axis after passing through the lens. Draw this ray passing through the focal point (F) on the left side of the lens.
b) The second ray passes through the optical center (O) undisturbed and continues in a straight line.
c) The third ray passes through the focal point (F) on the right side of the lens and refracts parallel to the principal axis after passing through the lens.
Step 6: After refracting, extend the rays on the opposite side of the lens. The point at which the rays intersect represents the location of the image. Mark this point.
Step 7: Draw an arrow from the marked point downward to represent the image. Measure the height of the image from the principal axis to this point.
By following these steps, you can locate the image produced by the diverging lens and determine its height.
To find the image distance and height using a diverging lens, we can use the lens formula and magnification formula.
The lens formula is:
1/f = 1/v - 1/u
Where:
f = focal length of the lens
v = image distance (the distance of the image from the lens)
u = object distance (the distance of the object from the lens)
In this case, the lens is a diverging lens with a focal length of 20 cm. The object distance (u) is given as 60 cm.
Substituting the values into the lens formula:
1/20 = 1/v - 1/60
Simplifying the equation:
3v - 2v = 60
v = 60 cm
So, the image distance (v) is 60 cm.
To determine the height of the image, we can use the magnification formula:
magnification (m) = -v/u
Using the values from the problem:
m = -60/60 = -1
Since the image is inverted because it is formed by a diverging lens, the magnification is negative.
The height of the image can be calculated using the magnification formula:
magnification (m) = height of the image (h') / height of the object (h)
Rearranging the formula:
h' = m * h
Substituting the magnification (m) value as -1 and given that the height of the object (h) is 10 cm, we have:
h' = -1 * 10 = -10 cm
So, the image is inverted and its height is 10 cm.