1) A mirror creates an image that has a height of 1.30 cm that is behind the mirror and 189 cm from the object.The image was created from a 2.60 cm tall object. What is the focal length of the mirror?
2) A lens placed 12.0 cm in front of an object creates an upright image 3.00 times the height of the object. The lens is then moved along the optical axis until it creates an inverted image 3.00 times the height of the object. How far did the lens move?
(I have been trying to figure these two out for a while, can someone please help me?)
Sure, I can help you solve these two problems!
1) To find the focal length of a mirror, we can use the mirror formula:
1/f = 1/di + 1/do
where f is the focal length, di is the image distance, and do is the object distance.
Given:
di = -189 cm (since the image is behind the mirror)
do = -2.60 cm (since the object is in front of the mirror)
Plugging in the values, we get:
1/f = 1/-189 + 1/-2.60
To simplify the equation, we take the reciprocal of both sides:
f = -2.60 * -189 / (-189 - -2.60)
Simplifying further, we get:
f ≈ 3.74 cm
Therefore, the focal length of the mirror is approximately 3.74 cm.
2) To find how far the lens moved, we need to determine the initial and final positions of the lens.
Given:
Initial position: object distance (do) = -12.0 cm
Image height (hi) / Object height (ho) = 3.00 (since the image is upright)
From the lens formula:
1/f = 1/do + 1/di
We can plug in the values for the initial position and solve for the initial focal length (f1):
1/f1 = 1/-12 + 1/di
Since the image is upright, di will also be negative.
Now, let's find the final position of the lens. The image is inverted, so the image height (hi) / object height (ho) will still be 3.00.
Using the lens formula again:
1/f2 = 1/do + 1/di
And replacing do with the final position (x) since the object distance changed:
1/f2 = 1/x + 1/di
Now, let's set up a ratio using the magnification equation:
hi/ho = -di/do
Given that the magnification (m) is 3.00:
3.00 = -di/do
Solving for di:
di = -3.00 * do
Substituting the value of di in terms of do in the equation for f2:
1/f2 = 1/x + 1/(-3.00 * do)
Since the object distance (do) remains unchanged, we can solve for x:
1/f2 = 1/x + 1/(-3.00 * do)
Now, we can find the difference in positions:
Difference = |x - (-12.0 cm)|
Simplify the equation and solve for the difference:
Difference ≈ |-36.0 cm - (-12.0 cm)|
Therefore, the lens moved approximately 24.0 cm along the optical axis.
I hope this helps! Let me know if you have any other questions.