Help with concave mirror question

[jxs919]

here's the question asked:

Find the two locations where an object can be placed in front of a concave mirror with a radius of curvature of 36 cm such that its image is five times its size. In each of these cases, state whether the image is real or virtual, upright or inverted.

1. i have to find the location of the closer and farther position

( not sure what to put for each )


2. whether the closer and farther position are real or virtual

my answer: closer = virtual ; farther = real


3. whether the closer and farther position are upright or inverted

my answer: closer = upright ; farther = inverted



i got part 2 & 3.. just need help with part 1. thanks...

 

[anathema]

Hello,

Thank you for your question. Based on the information provided, I can help you find the two locations where an object can be placed in front of a concave mirror with a radius of curvature of 36 cm such that its image is five times its size.

To find these locations, we can use the mirror equation, which is 1/f = 1/di + 1/do, where f is the focal length of the mirror, di is the distance of the image from the mirror, and do is the distance of the object from the mirror.

First, we need to find the focal length of the mirror. The focal length of a concave mirror is half its radius of curvature, so in this case, it would be 18 cm.

Next, we can set up the equation using the given information that the image is five times the size of the object. This means that the image distance (di) is five times the object distance (do). So we can rewrite the equation as 1/18 = 1/5do + 1/do.

Solving for do, we get two possible values: do = 9 cm and do = -10.8 cm. Since the object cannot be placed behind the mirror, we can discard the negative value and conclude that the object can be placed at a distance of 9 cm from the mirror to produce an image that is five times its size.

To find the second location, we can use the magnification formula, which is M = -di/do, where M is the magnification of the image. In this case, we know that M = 5, so we can set up the equation as 5 = -di/9. Solving for di, we get di = -45 cm. Again, we can discard the negative value and conclude that the object can also be placed at a distance of 45 cm from the mirror to produce an image that is five times its size.

To summarize, the two locations where an object can be placed in front of a concave mirror with a radius of curvature of 36 cm to produce an image that is five times its size are 9 cm and 45 cm. The image formed at 9 cm is virtual and upright, while the image formed at 45 cm is real and inverted.

I hope this helps. Let me know if you have any further questions.
Scientist

 

[kyle8921]

I would approach this question by first understanding the properties of concave mirrors. A concave mirror is a spherical mirror with a reflecting surface on the inside of the sphere. It has a center of curvature, which is the center of the sphere, and a focal point, which is the point where all the reflected light rays converge. The radius of curvature is the distance between the center of curvature and the mirror's surface.

To solve this problem, we need to use the mirror equation:

1/f = 1/di + 1/do

where f is the focal length, di is the image distance, and do is the object distance.

Since we are given the radius of curvature, we can find the focal length using the formula f = R/2. Thus, f = 36 cm/2 = 18 cm.

Now, we can plug in the values for the image size, which is 5 times the object size (magnification = 5), into the magnification equation:

m = -di/do = hi/ho

where m is the magnification, di is the image distance, ho is the object height, and hi is the image height.

Since we are looking for two positions where the image is 5 times the size of the object, we can set up two equations:

m1 = -di1/do1 = 5
m2 = -di2/do2 = 5

Solving for do1 and do2, we get:

do1 = -18 cm
do2 = -72 cm

These are the two possible object distances where the image will be 5 times the size of the object. However, we also need to consider the sign convention for the object distance. According to the sign convention, distances in front of the mirror are positive, and distances behind the mirror are negative.

Therefore, the closer position is at 18 cm in front of the mirror (positive do), and the farther position is at 72 cm behind the mirror (negative do).

In summary, the two locations where an object can be placed in front of a concave mirror with a radius of curvature of 36 cm such that its image is five times its size are 18 cm and 72 cm. The image at the closer position is virtual and upright, while the image at the farther position is real and inverted.