# Reflection questions help?

1. Nov 12, 2012

### skg94

1. The problem statement, all variables and given/known data

Im having problems with these questions i just cant seem to get them
Any help would be good thank you in advance

1. An object located 40cm in front of a mirror produces an erect image 80cm from the mirror. Find the radius of curvature for the mirror, and state the type of mirror. (ans:160cm,concave)

2. An object located 40cm infront of a mirror produces an inverted image 120cm from the mirror. Find the radius of curvature for the mirror, and state the type of mirror. (ans:60cm,concave)

3. An object located 40cm infront of a mirror produces an erect image 20cm from the mirror. Find the radius of curvature for the mirror, and state the type of mirror. (-80cm,convex)

4. A 20cm object located 30cm in front of a mirror generates an erect image that is 10 cm tall. What is the size of image produced when he object is moved 60cm further from the mirror's surface? (5cm)

5. An object located infront ofa concave mirror with a radius of curvature of 80cm produced an inverted image that is three times the size of the object. What is the object distance? (53)

6. An object located infront of a concave mirror with a radius of curvature of 180cm produced an erect image that is two times the szie of the object. What is the object distance? (45)

7. An object located in front of a convex mirror with a focal length of 60cm produced an erect image that is 1/6 the size of the object. What is the object distance? (300cm)

2. Relevant equations

1/f=1/d(object)+1/d(image)

m=Hi/Ho=-Di/Do
radius = c = 2f

3. The attempt at a solution

1 f= (do^-1+di^-1)^-1
= (40^-1) +(80^-1) ^ -1 = 26.6
c = 53 , which is wrong.

For the next 3 questions i substitued the values as i did above, perphaps my equation is wrong or am i substituting wrong?

Question 4, since its moved to 60cm which is double the previous, does that make it just half the image height based on ratio?

5-7 i have no idea

2. Nov 12, 2012

### Simon Bridge

If the equation does not work for you - try sketching the ray diagram and use normal geometry.
You can get a long way through these problems just by understanding the way mirrors work.
... so the image is twice as far away as the object and erect.

What kind if image in a mirror is upright? Real or virtual?
Is the image in front or behind the mirror?

Draw a line for the mirror, draw a short arrow for the object some distance in front of the mirror.
Draw a line for the image location. Draw the ray from the object to the center of the mirror ... and continue it, with the appropriate reflection rule, to the image line.
Draw a horizontal line from the object to the mirror ... and from there to where the last line met the image line: that will tell you the reflection angle and, thus, the location of the focus.

Where, in terms of the radius of curvature, does the object have to be, for the image to be twice as far away?

Last edited: Nov 12, 2012
3. Nov 12, 2012

### skg94

4. Nov 12, 2012

### Simon Bridge

I'm afraid it is not clear what your statement refers to... because you have quoted several possibilities.
Try to identify where you get lost.

I'll assume you know what a ray diagram is and you've heard of geometry ... so you must understand the first statement to the "...."
If the image is at 40cm and the image at 80cm then the image is twice as far as the object right?
So it must be the questions...

Lets take those questions one at a time:
Do you not understand what a real or a virtual image is?

The statement says that the image is upright ... it is caused by a single curved mirror.
From that information: is that image real or is it virtual.

5. Nov 13, 2012

### skg94

these questions arent theoretical but mathematical the location does really matter i believe its the application of the formula, they all say they are infront so they are all real, i just cannot get the radius of curvature using the formulas mathematically

6. Nov 13, 2012

### Simon Bridge

Makes no difference to how they are solved
You have to get the sign convention correct to use the formula - which means, for one thing, correctly identifying whether the image is real or virtual. Then the formula will tell you the focal length (which tells you the radius of curvature) and the sign of the focal length tells you the type of mirror.

But you are having trouble with the formula - you can sort out the problem by sketching the ray diagram. In fact - when you do it, the answer will be obvious.
No they don't - the first question for example just tells you that the image is 80cm away from the mirror - it does not say if that is in front or behind the mirror.

Check: Is it even possible for an upright image from a mirror to be real?
The radius of curvature of a spherical mirror has a special relationship with the focal length.

What these questions are doing is testing your understanding of how mirrors work.
That's why I'm not just giving you the answers ... instead, I am asking you questions that will guide you to the right answers. If you will not answer these questions, I cannot help you. If you will not sketch the diagram, I cannot help you.

7. Nov 13, 2012

### skg94

oh i see where your coming from, for the first the image is virtual, which works applied to the formula, it is possible, an image inbetween the focus and vertex is an upright, bigger image if im correct?

drawing the ray diagrams helped me alot for the first 3 and i got them all, thank you for suggesting that just didnt come to me.

Can you help me with the ratio problems? 5-7 im assuming they are ration problems

8. Nov 13, 2012

### Simon Bridge

Where does the object have to be to make that image?
I had a look at your other questions and I see you prefer to use equations ... the diagrams are usually more reliable and there is less to memorize ;)
OK - lets see:

You can still use the ray diagrams :)

1. is this image real or imaginary?
2. draw the mirror and principle axis and focus like usual
3. draw a horizontal ray 1 unit above the principle axis - this is the height of the object
4. draw a horizontal ray 3 units below the principle axis - this is the height of the image
5. reflect the above two rays in the mirror - where the first one crosses the the second one is the image position and where the second crosses the first is the object position.
Cool huh?

Using equations - how would you find the magnification?

9. Nov 13, 2012

### skg94

Actually i have a question about the ray diagrama when it doesnt specify where the object is on the principle where do i draw the obect?
Magniication is the hi/ho=-di/do

10. Nov 13, 2012

### Simon Bridge

You don't need to draw the object in to answer the question - you only need to know where the horizontal ray from the object hits the mirror. (BTW: you draw a straight, vertical, line for the mirror right - not a curve? All these optics use an approximation that the radius of the mirror is much smaller than it's radius of curvature.)

You can still find it if you want - for completeness. The basic approach is the same as for finding the image position ... see step 5 above. There are three principle rays, and two of them give you horizontal lines. The horizontal ones get reflected through the focus ... these rays will cross in two places.

From the equation approach - you know M so $d_i=-Md_0$ ... and you have another equation that relates these two distances to the focal length. Two equations and two unknowns...