Should Approved Sign Conventions be Applied in Optics Problems Using Lens and Mirror Equations?

[jishnu]

We have the mirror equation
(1/v) +(1/u)=(1/f)
and the lens equation
(1/v) - (1/u)=(1/f)
Where
u =object distance from the mirror/lens
v =image distance from the mirror/lens
f =focal length of the mirror /lens
My doubt is, when we are doing some problems in optics using these equations should we to apply the approved sign conventions to these physical values ie;
Standard sign conventions:
the distance measured to the left and bottom from the pole of the lens/mirror is - ve and distance measured to the right and top from the pole of the lens is +ve.

 

[kuruman]

The equation I know is the same for both mirrors and thin lenses, namely
1/v + 1/u = 1/f
where
u =object distance from the mirror/lens
v =image distance from the mirror/lens
f =focal length of the mirror /lens
Now u and v are distances from the object and image to the lens on the side where "light has every business being". This means that u and v are on the same side in the case of a mirror and opposite sides in the case of a lens. With this convention, a negative v signifies a virtual image because it appears where "light has no business being."
I am not sure what you mean by "standard" conventions because your lens equation with its negative sign is not standard. Also, I am not sure what you mean by "the distance measured to the left and bottom from the pole of the lens/mirror".

 

[Merlin3189]

Have a look at Hyperphysics. They have a brief reference to both the common Gaussian form and the more esoteric Cartesian form. No doubt you have more advanced texts that go into more detail.

For the Gaussian form that we all learn first, there is more than one convention. I learned the "Real is Positive", but I think I've seen others in use on PF. I hadn't come across your Cartesian convention until I saw this post.

 

[jishnu]

The equation I know is the same for both mirrors and thin lenses, namely
1/v + 1/u = 1/f
where
u =object distance from the mirror/lens
v =image distance from the mirror/lens
f =focal length of the mirror /lens
Now u and v are distances from the object and image to the lens on the side where "light has every business being". This means that u and v are on the same side in the case of a mirror and opposite sides in the case of a lens. With this convention, a negative v signifies a virtual image because it appears where "light has no business being."
I am not sure what you mean by "standard" conventions because your lens equation with its negative sign is not standard. Also, I am not sure what you mean by "the distance measured to the left and bottom from the pole of the lens/mirror".

It may be my mistake that I have used the word "standard", I shall replace it by "generally used sign convention (attached a reference from my textbook regarding that)
I too got confused when using these different equations for mirror and lens, that's what made me ask about that
But, the textbook has given separate formula for mirror and lens(photo of the summary from the textbook is attached within this post)
There we can see the two formulas given separately.

 

[kuruman]

You should use the convention that you learned and you are most comfortable with. Your textbook uses the Cartesian convention which views the mirror/lens as the origin. Then the convention "to the left" negative - "to the right" positive is observed. Both the Gaussian and Cartesian forms should give the same result if you use them correctly and interpret the results correctly. Consider the following problem: An object is placed at distance u = 3f from a lens of focal length f to the left of the lens. Find the position v of the image in units of f.
Solution
Gaussian way
1/u + 1/v = 1/f. Here u = 3f,
1/(3f) + 1/u = 1/f → 1/u = 1/f - 1/(3f) = 2/(3f)
∴ u = (3/2)f
Interpretation
Because u is positive, the image is real and appears to the right of the lens at distance (3/2)f.
Cartesian way
1/v - 1/u = 1/f Here u = -3f because the object is to the left of the lens.
1/v - 1/(-3f) = 1/f → 1/v + 1/(3f) = 1/f
This last equation is the same as the Gaussian equation and its solution proceeds as before and u = (3/2)f.
Interpretation
The answer came out positive therefore the image appears to the right of the lens.

 

[jishnu]

You should use the convention that you learned and you are most comfortable with. Your textbook uses the Cartesian convention which views the mirror/lens as the origin. Then the convention "to the left" negative - "to the right" positive is observed. Both the Gaussian and Cartesian forms should give the same result if you use them correctly and interpret the results correctly. Consider the following problem: An object is placed at distance u = 3f from a lens of focal length f to the left of the lens. Find the position v of the image in units of f.
Solution
Gaussian way
1/u + 1/v = 1/f. Here u = 3f,
1/(3f) + 1/u = 1/f → 1/u = 1/f - 1/(3f) = 2/(3f)
∴ u = (3/2)f
Interpretation
Because u is positive, the image is real and appears to the right of the lens at distance (3/2)f.
Cartesian way
1/v - 1/u = 1/f Here u = -3f because the object is to the left of the lens.
1/v - 1/(-3f) = 1/f → 1/v + 1/(3f) = 1/f
This last equation is the same as the Gaussian equation and its solution proceeds as before and u = (3/2)f.
Interpretation
The answer came out positive therefore the image appears to the right of the lens.

In that way it is ohk, but can you solve the same question for mirrors? Also what about the sign convention for their focal length?

 

[kuruman]

In that way it is ohk, but can you solve the same question for mirrors? Also what about the sign convention for their focal length?

In the Gaussian method the same equation is used for mirrors and lenses. See post #2. The focal lengths of convex mirrors and diverging lenses that always produce virtual images are negative.
Problem
An object is placed at 2f to the left of a diverging length of focal length 10 cm. Find the position of the image.
Gaussian way
1/u + 1/v = 1/f → 1/v = 1/f - 1/u
1/v = 1/(-10) - 1/20 = -3/20
∴ v = -20/3 = -6.7 cm
Interpretation
The position of the image is negative which means that it is virtual, i.e. it appears on the same side as the object, at 6.7 cm from the lens, where light has no business being.
Cartesian way
1/v - 1/u = 1/f → 1/v = 1/f - 1/u
1/v = 1/(-10) - 1/20 = -3/20
∴ v = -6.7 cm
Interpretation
The position of the image is negative which means that it appears on the same side as the object, at 6.7 cm from the lens, where light has no business being. Therefore it is virtual.

 

[jishnu]

Cartesian way
1/v - 1/u = 1/f → 1/v = 1/f - 1/u
1/v = 1/(-10) - 1/20 = -3/20
∴ u = -6.7 cm
In this case how can we use that equation according to the photo of the text summary we have to use the equation
1/v +1/u = 1/f
So, according to the questions condition:
u = - 20 cm
f = - 10 cm
Therefore,
1/v =1/f - 1/u
= 1/(-10) - 1/(-20)
= 1/20 - 1/10
= - 1/20
v = - 20
What's wrong with this method, but this gives a totally different answer.
How come that happens?

 

[kuruman]

In this case how can we use that equation according to the photo of the text summary we have to use the equation ...

Please state clearly what problem you are trying to solve and what method you propose to use. Use the following format, specify the optical instrument and fill in the blanks with the appropriate numbers.

An object is placed at _______ cm to the left of a ______________ (concave/convex or converging/diverging) ______________ (mirror or lens) of focal length _______ cm. Find the position of the image using the ___________ (Gaussian or Cartesian) method.

 

[jishnu]

Please state clearly what problem you are trying to solve and what method you propose to use. Use the following format, specify the optical instrument and fill in the blanks with the appropriate numbers.

An object is placed at _______ cm to the left of a ______________ (concave/convex or converging/diverging) ______________ (mirror or lens) of focal length _______ cm. Find the position of the image using the ___________ (Gaussian or Cartesian) method.

Its the exact same question that you have solved in the post #7
I tried solving it using the mirror equation given in my textbook (photo is already posted) in the Cartesian method.

 

[kuruman]

Post #7 is about a lens (misspelled length) not a mirror. Anyway, let's do it for a concave mirror.

Problem
An object is placed at 2f to the left of a concave mirror of focal length 10 cm. Find the position of the image.
Gaussian method
1/u + 1/v = 1/f → 1/v = 1/f - 1/u
1/v = 1/(10) - 1/20 = 1/20
∴ v = 20 cm
Interpretation
The position of the image is positive which means that it is real, i.e. it appears on the same side as the object, at 20 cm from the mirror, where light has every business being.
Cartesian method
1/u + 1/v = 1/f → 1/v = 1/f - 1/u
1/v = 1/(-10) - 1/(-20) = -1/20
∴ v = -20 cm
Note: Your book says that for a convex mirror, f is positive. Here we have a concave mirror, which means f = - 10 cm.
Interpretation
The position of the image is negative which means that it appears on the same side as the object (to the left of the mirror), at 20 cm from the mirror, where light has every business being. Therefore it is real.

 

[jishnu]

Post #7 is about a lens (misspelled length) not a mirror. Anyway, let's do it for a concave mirror.

Problem
An object is placed at 2f to the left of a concave mirror of focal length 10 cm. Find the position of the image.
Gaussian method
1/u + 1/v = 1/f → 1/v = 1/f - 1/u
1/v = 1/(10) - 1/20 = 1/20
∴ v = 20 cm
Interpretation
The position of the image is positive which means that it is real, i.e. it appears on the same side as the object, at 20 cm from the mirror, where light has every business being.
Cartesian method
1/u + 1/v = 1/f → 1/v = 1/f - 1/u
1/v = 1/(-10) - 1/(-20) = -1/20
∴ v = -20 cm
Note: Your book says that for a convex mirror, f is positive. Here we have a concave mirror, which means f = - 10 cm.
Interpretation
The position of the image is negative which means that it appears on the same side as the object (to the left of the mirror), at 20 cm from the mirror, where light has every business being. Therefore it is real.

Thanks allot sir, now its clear for me. So both the methods provides us the same answer.
Where can I get more information about the Guassian method because I feel it more easy and logical than the method provided in my textbook.

 

[kuruman]

Thanks allot sir, now its clear for me. So both the methods provides us the same answer.
Where can I get more information about the Guassian method because I feel it more easy and logical than the method provided in my textbook.

I agree with you that the Gaussian method is easier to understand and learn. You can find how to use it here for lenses
https://www.khanacademy.org/science...nses/v/thin-lens-equation-and-problem-solving
and a summary of the three methods (there is also the Newtonian method) here
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

You can do your own internet search for "thin lens equation" and I am sure you will find many solved problems in PF threads. Likewise for "spherical mirror".

 

[jishnu]

I agree with you that the Gaussian method is easier to understand and learn. You can find how to use it here for lenses
https://www.khanacademy.org/science...nses/v/thin-lens-equation-and-problem-solving
and a summary of the three methods (there is also the Newtonian method) here
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

You can do your own internet search for "thin lens equation" and I am sure you will find many solved problems in PF threads. Likewise for "spherical mirror".

Thanks allot sir.[emoji4]