Mirrors and Lenses equation

In summary: Ho is positive as it is an upright object. In the equation Hi/Ho = -Di/Do, the minus sign is there to make this true. You can also think of the minus sign as telling you that Di and Ho have opposite signs. Ho is positive; therefore, Di is negative.The minus sign is also in the equation 1/f = 1/Di + 1/Do. It is there because Di and Do have opposite signs, as indicated by their respective subscripts.When I do the same with the convex lense I get the same answer but the sign is switched in the final answer.If the there was no negative # in 1/(1/3Do) then I will get 4/
  • #1
mikil100
12
0

Homework Statement


In front of a spherical concave mirror of radius 39cm, you position and object of height 0.6cm somewhere along the principal axis. The resultant image has a height of 0.2cm, How far from the mirror is the object located?
What if this were a convex mirror with the same radius, and the same image and object height?

Homework Equations



1/f=1/Di+1/Do

M=Hi/Ho= -Di/Do

F=1/2R

The Attempt at a Solution



1/2R=19.5cm focal point

I have two equations with two unknowns (Di and Do). I used the equation Hi/Ho=-Di/Do to solve the equation for -Di, getting 0.2/0.6= 1/3. 1/3=-Di/Do, solving for Di I get -1/3Do=Di.

I then just plug this equation into the mirror equation

1/do+1/(-1/3do)=1/f... I bring the -3 from the denominator into the numerator

1/f is 2/39

1/d0-3/do=2/39

-2/do=2/39
do=-39cm

I follow the same process for a convex mirror but I get a positive answer with the same value of 39. I don't think a negative answer makes sense, and I am unsure of my sign usage... I find it odd that my object distance is the same as my mirrors radius.

Thank you in advance for the help.
 
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  • #2
mikil100 said:
1/do+1/(-1/3do)=1/f
Where does the minus sign come from?

mikil100 said:
do=-39cm
This would mean the object is right in the center of the sphere (or on the wrong side, depending on the sign convention). Can that be true?
 
  • #3
mikil100 said:

Homework Equations



1/f=1/Di+1/Do

M=Hi/Ho= -Di/Do

F=1/2R

The Attempt at a Solution



1/2R=19.5cm focal point

I have two equations with two unknowns (Di and Do). I used the equation Hi/Ho=-Di/Do to solve the equation for -Di, getting 0.2/0.6= 1/3. 1/3=-Di/Do,

The image is smaller than the object. In case of a concave mirror, that means real image, which is upside-down, so Hi is negative, and Hi/Ho = -1/3 = -Di/Do.
 
  • #4
mfb said:
Where does the minus sign come from?

This would mean the object is right in the center of the sphere (or on the wrong side, depending on the sign convention). Can that be true?

Hi, thank you for the reply.

The negative sign in 1/(-1/3Do) is from this equation M=Hi/Ho= -Di/Do

Where Di is negative, getting Di on its own I have 0.2/0.6=-Di/Do----> 1/3*Do=-Di multiply both sides by -1 to get -1/3Do=Di

I then plugged the negative 1/3Do into 1/Di in the mirror equation. This negative sign is what is really tripping me up actually. When I do the same with the convex lense I get the same answer but the sign is switched in the final answer.

If the there was no negative # in 1/(1/3Do) then I will get 4/Do=1/F, which equals 78cm.
 
  • #5
.
mikil100 said:
The negative sign in 1/(-1/3Do) is from this equation M=Hi/Ho= -Di/Do

Read my Post #3 . Hi is negative as it is an upside-down image.
 

What is the equation for mirrors and lenses?

The equation for mirrors and lenses is known as the thin lens equation, which is:

1/f = 1/do + 1/di

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

How is the thin lens equation derived?

The thin lens equation is derived using the laws of refraction and the geometry of a thin lens. It is based on the principle that light rays passing through a thin lens will converge or diverge depending on the curvature of the lens and the distance of the object and image from the lens.

What is the significance of the thin lens equation?

The thin lens equation is significant because it allows us to calculate the location and size of an image formed by a lens, given the object distance and focal length. This is useful in understanding how lenses work and in designing optical systems.

Can the thin lens equation be applied to all types of lenses?

Yes, the thin lens equation can be applied to all types of lenses, including convex, concave, and even complex lenses made up of multiple elements. However, it assumes that the lenses are thin, meaning their thickness is negligible compared to their focal length.

What are the limitations of the thin lens equation?

The thin lens equation is based on several assumptions, such as thin lenses, parallel rays of light, and negligible aberrations. In reality, these assumptions may not hold true, especially for more complex lenses. As a result, the thin lens equation may not always give accurate results and should be used with caution.

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