a) The lens is converging, because the object is placed in front of the lens and the image is located behind the lens.
b) The focal length of the lens is 60 mm.
c) The diagram would look like this:
Object: x = -30 mm
Lens: x = 0 mm
Image: x = 90 mm
d) The image is real, smaller than the object, and inverted in comparison to the object.
e) The diagram would look like this:
Object: x = -30 mm
Mirror: x = 0 mm
Image: x = -20 mm
f) The height of the image is 7.5 mm and the magnification of the mirror is 0.5.
An object is placed 30mm in front of a lens. An image of the object is located 90mm behind the lens.
a) Is the lens converging or diverging? explain.
b) What is the focal length of the lens?
c) Draw a diagram with lens at x=0, locate the image.
d) Is the image real or virtual?
Is it smaller, larger, or same size as the object?
Is it inverted or upright in comparison to the object?
e)the lens is replaced by a concave mirror with focal lenght 20 mm. Draw the mirror at position x=0 so that a real image is formed. Draw at least two rays and locate the image to show this situation.
f) If the object height is 15 mm, calculate the height of the image and the magnification of the mirror.
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