Light, Mirrors, Lenses, immediate help required

AI Thread Summary
The discussion revolves around a physics problem involving a concave mirror, where an object is placed 12 cm from the mirror and the goal is to find the image location. The user initially struggles with applying the mirror equation and magnification concepts. A key point made is that the focal length can be determined using the sun's position, which acts as an object forming an image 3 cm away. By utilizing the magnification equation, it is concluded that the object should be placed 24 cm from the mirror to achieve the desired image formation. This solution provides clarity for the user as they prepare for their assignment.
mhl2188
Messages
1
Reaction score
0
Hello, here is the problem i am having trouble with. The sun falls on a concave mirror and forms an image 3 cm from the mirror. If an object 24mm high is placed 12 cm from the mirror, where will its image be formed? So far, i have decided that since 1/f = 1/do + 1/di...
1/f = 1/do + 1/3 = 1/12 + 1/di
I also decided that hi/2.4 = -di/12... The problem is...even if i solve for di in terms of hi and plug that into my first equation i don't have enough information to solve for do. I am stuck and this problem is due tomorrow... please help
thanks
mhl
 
Physics news on Phys.org
mhl2188 said:
Hello, here is the problem i am having trouble with. The sun falls on a concave mirror and forms an image 3 cm from the mirror. If an object 24mm high is placed 12 cm from the mirror, where will its image be formed? So far, i have decided that since 1/f = 1/do + 1/di...
1/f = 1/do + 1/3 = 1/12 + 1/di
I also decided that hi/2.4 = -di/12... The problem is...even if i solve for di in terms of hi and plug that into my first equation i don't have enough information to solve for do. I am stuck and this problem is due tomorrow... please help
thanks
mhl

You need to use the "sun" information to find the focal length of the mirror. The sun is an object at distance ___________ that forms an image at 3cm. What is f?
 


Hello mhl,

Thank you for reaching out for help with your problem. It seems like you have made some progress in understanding the problem and utilizing the lens/mirror equation 1/f = 1/do + 1/di. However, you are correct in realizing that you do not have enough information to solve for do.

To solve this problem, we need to use the magnification equation, m = -di/do = hi/ho. This equation relates the height of the image (hi) to the height of the object (ho) and the distances of the image (di) and object (do) from the mirror or lens.

Since we know the height of the object (24mm) and the height of the image (3cm = 30mm), we can set up the equation as follows:

m = -di/do = hi/ho
m = -30/do = 30/24
m = -5/4

Now, we can use this value of m to solve for do:

m = -di/do
-5/4 = -30/do
do = 24 cm

Therefore, the object must be placed 24 cm away from the concave mirror in order to form an image 3 cm away from the mirror. I hope this helps and good luck with your assignment!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top