How to Use the Mirror Equation to Solve Convex & Concave Problems in Optics

[mustang]

Problem 8.
A concave amkeup mirror is designed so that a person 28.7 cm in front of it sees an upright image ata distance of 51.1 cm behind the mirror.
What is the radius of curvature of the mirror? Answer in cm.
Note: What is the radius of curvature formula?

Problem 19.
A convex mirror of focal length 32 cm forms an image of a soda bottle at a distance of 19 cm behind the mirror. The height of the image is 7 cm.
Where is the image located? In units of cm.
Note: Where do I start?

Problem 23.
A spherical glass ornament is 7.44 cm in diameter. An object is placed 9.62 cm away from the ornament.
Where will its image form? In units of cm.
Note: What formula should I use?

Problem 32.
A concave spherical mirror can be used to project an image onto a sheet of paper, allowing the magnified image of an ulluminated real object to be accurately traced.
If you have a concave mirror with a focal length of 16 cm, where would you place a sheet of paper so that the image projected onto it is twice as far from the mirror as the object is? In units of cm.
Note: I don't know where to start/

Problem 38.
A convex mirror with a radius of curvature of 45.9 cm forms a 1.71 cm tall image of a pencil at a distance of 15 cm behind the mirror.
What is the height of the object? In units of cm.
Note: I don't know where to start!

 

[Severian596]

Problem 8:
(see http://www.sasked.gov.sk.ca/docs/physics/u3b32phy.html )
According to this site, the curved mirror and lens equation (Gaussian form) is:

$$1/d_{o}+1/d_{i}=2/R$$

where $$d_{o}$$ is the distance from the object to the vertex (mirror), $$d_{i}$$ is the distance from the image to the vertex (mirror), and finally $$R$$ is the radius of curvature.

Keep in mind that an image's focal length $$f$$ is related to the radius of curvature according to:

$$R = 2f$$

 

[Severian596]

Originally posted by mustang
Problem 19.
A convex mirror of focal length 32 cm forms an image of a soda bottle at a distance of 19 cm behind the mirror. The height of the image is 7 cm.
Where is the image located? In units of cm.
Note: Where do I start?

I believe you have a typo in this problem. I'll assume the question is "Where is the OBJECT located?"

You are given the Focal Length $$f=32$$ and the distance from the image to the mirror $$d_{i} = 19$$ . Use the equation

$$1/d_{o}+1/d_{i}=2/R$$

and substitute

$$R = 2f$$

to find $$d_{o}$$ .

 

[Severian596]

Finally, I recommend three things so you can find the answers to the rest of these problems:

1) Review some of the webpage I mentioned above.
2) Review this web page for more info (and the sections it's linked to up the left side): http://www.glenbrook.k12.il.us/gbssci/phys/Class/refln/u13l3a.html
3) Read your text.

Your questions smack of laziness. I'm not saying this IS the issue, but in 10 minutes of internet research I found two sites with leads the the answers to your questions. I assume you have a textbook and that it can offer some assistance, too.

 

[mustang]

Problem 8.

So would it be 1/28.7+1/51.1 =2/R or
0.03483+0.019569=2/R
0.054399=2/R
0.054399(1/2)=R
0.027199=R

 

[Severian596]

Originally posted by mustang
So would it be 1/28.7+1/51.1 =2/R or
0.03483+0.019569=2/R
0.054399=2/R
0.054399(1/2)=R
0.027199=R

Be careful of your algebra, mustang. You made a rather costly mistake between these two steps:

0.054399=2/R
0.054399(1/2)=R

Once you divide by 2 on both sides, the R is still 1/R.

$$(0.054399)(1/2)=1/R$$

then solving for R, you'll get

$$R = 2/0.054399$$

I was leery right off the bat when I saw that your answer for the radius of the mirror was a fraction of a cm! I'm sure you were, too.

 

[mustang]

The answer was wrong ):'

I did 2/0.054399=36.7653817 and that was wrong?

 

[Severian596]

Based on the correct answer find out what went wrong. I didn't mislead you and the equation was right there...perhaps something was lost in translation?

 

[mustang]

In Regards to problem 32.

To start solving this problem I did 1/8.5=2/-q+1/q
then, 1/8.5=-1/q
q=-8.5
Is this right? I feeling that it isn't!

problem31.
A concave shaving mirror has a radius of curvature of 23.4 cm. Find the magnification of the image when an upright pencil is placed 6.11 cm from the mirror. Describe the image.
First, to find magnification: p=6.11
23.4=2f
11.7=f
So 11.7^-1-6.11^-1=ans^-1=-12.78837209.
Now for magnification:
M=-q/p
-(-12.78837209)/6.11=2.093023256


So to describe the image. Since i need to indicate if the image is virtual/real, smaller/larger, and upright/inverted? I got virtual upright larger. Is this right?

 

[mustang]

On Problem 32 help! Due midnight!

I got f=11cm.
I figured out that the formula I should use is 1/f=1/p+1/q
so would it be 1/f=2/-q+1/q
If so i need help to go on if not i need more help please this is due by Midnight.