An object, O, is placed 18 cm from the center of a converging lens of focal length, f= 6 cm as shown.
O F F
(a) On the illustration, sketch a ray diagram to locate the image.
(b) Is the image real or virtual? Explain your answer.
(c) Using the lens equations, compute the image distance.
(d) Calculate the Height of the image.
There is no illustration and we don't draw sketches.
I assume you know the lens equation, so use it to do part (c)
1/Do + 1/Di = 1/f
The image is real if Di, the image distance, is positive
For (d), the magnification is Do/Di
(a) To sketch a ray diagram to locate the image, we need to consider two different rays of light coming from the object O. The first ray should be parallel to the principal axis and the second ray should pass through the focal point F.
Start by drawing a horizontal line representing the principal axis of the lens. Place the object O on the left side of the lens, 18 cm away from the center. Draw a straight line from the top of the object horizontally until it intersects with the principal axis. This represents the first ray.
Next, draw another straight line from the top of the object through the focal point F of the lens. This represents the second ray.
Finally, draw a converging lens shape with the principal axis passing through its center, with the focal point F on one side and the object O on the other side. The intersection point of the two rays will determine the location of the image.
(b) To determine if the image is real or virtual, we need to analyze the direction of the rays after passing through the lens. In this case, the rays converge on the same side of the lens as the object, which means they do not actually intersect. Thus, the image is virtual.
(c) To compute the image distance using the lens equation, we can use the formula:
1/f = 1/o + 1/i
where f is the focal length, o is the object distance, and i is the image distance.
Given:
f = 6 cm (focal length)
o = 18 cm (distance of the object from the center of the lens)
Plugging in the values, we can solve for i:
1/6 = 1/18 + 1/i
Simplifying the equation:
1/6 = (18 + i)/18i
Cross-multiplying:
6(18i + i) = 18i
108i + 6i = 18
114i = 18
i = 18/114
Therefore, the image distance is approximately 0.158 cm.
(d) To calculate the height of the image, we can use the magnification formula:
magnification = height of image / height of object = -image distance / object distance
Given:
i = 0.158 cm (image distance)
o = 18 cm (object distance)
Plugging in the values, we can solve for the height of the image:
magnification = -0.158 cm / 18 cm
Since the magnification is not given, we cannot directly compute the height of the image. However, if we know the height of the object, we can multiply it by the magnification to find the height of the image.