What Formula Do I Need for This Optics Problem?

[ohhwhat]

Homework Statement

The distance from a 6 cm high object to a screen is 80 cm. Where should a lens with a 15 cm focal length be set up so that it creates a sharp (enlarged or reduced) image of an object on the screen? How big are the pictures in both cases?

Relevant Equations

none

Unfortunately, there's no formula i know about, so i have no idea how to solve this...
Hope you can help me out...

Greetings

 

[etotheipi]

You may use the optic equation $\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$. Here $u$ is the position of the object in front of the lens and $v$ is the position of the image behind the lens. For a sharp image to form on the screen, the screen must be positioned where the image is formed.

There is a constraint of the form $f(u,v) = 0$ that allows you to eliminate either $u$ or $v$. What is it?

 

[ohhwhat]

So the distance of 80cm is the result of u + v, am i right?
Also, the solution we are looking for is u, right?

 

[etotheipi]

That is the constraint, yes. Now find the solutions for $u$.

 

[ohhwhat]

Mhm, what should i do with the heigh of the object (6cm)? :/
Also, i have either u or v, so how should i work with the optic equation right now?

 

[etotheipi]

Mhm, what should i do with the heigh of the object (6cm)? :/

You will use this to determine the sizes of the images. By considering similar triangles you will see that the ratio of the image size to the object size, $H/h$, equals the ratio $v/u$.

Also, i have either u or v, so how should i work with the optic equation right now?

Write $v = 80 - u$, and express the optic equation in terms of $u$ and $f$. Note that $f$ is given in the question, allowing you to find the solutions for $u$.

 

[ohhwhat]

So, u+v=80cm
Maybe i can use v=80−u in some way?
I struggled half an hour to solve it now, unfortunately, i didn't came to a solution...

 

[etotheipi]

Do you see \begin{align*}

\frac{1}{u} + \frac{1}{v} &= \frac{1}{f} \\ \\
\implies \frac{1}{u} + \frac{1}{80-u} &= \frac{1}{f}

\end{align*}$f$ is a constant. Do you know how to solve this equation?

 

[ohhwhat]

Yeah i had the same idea like you, unfortunately i don't know how to solve that.^
Maybe this: 1/f= 2/80-2u ?

 

[etotheipi]

I suggest you write it as
\begin{align*}

\frac{1}{u} + \frac{1}{80-u} & = \frac{1}{f} \\ \\

\implies \frac{1}{u} \cdot \underbrace{\left(\frac{80-u}{80-u}\right)}_{=1} + \frac{1}{80-u} \cdot \underbrace{\left(\frac{u}{u}\right)}_{=1} &= \frac{1}{f} \\ \\

\implies \frac{80-u}{u(80-u)} + \frac{u}{u(80-u)} &= \frac{1}{f}

\end{align*}Do you see why? We re-wrote each fraction in terms of a common denominator, by sneakily multiplying each fraction by $1$ (which doesn't change the fraction).

Can you now simplify that last line, and then solve the equation for $u$?

 

[ohhwhat]

No, sorry, i don't really understand this

 

[etotheipi]

You may need to review how to add fractions. What is, for instance, $\dfrac{1}{a} + \dfrac{1}{b}$?

 

[ohhwhat]

Maybe it's 2/a+b?

 

[etotheipi]

Maybe it's 2/a+b?

Maybe, but no

The rule for adding fractions is not $\dfrac{w}{x} + \dfrac{y}{z} = \dfrac{w + y}{x+z}$, but instead $\dfrac{x}{z} + \dfrac{y}{z} = \dfrac{x+y}{z}$. In other words, you first need to re-write the fractions so that they have the same denominator, and then you may add the numerators over this common denominator. For instance,$$\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{b}{ab} + \dfrac{a}{ab} = \dfrac{b+a}{ab}$$I would advise that you briefly review this section of algebra before returning to the optics question.

 

[berkeman]

No, sorry, i don't really understand this

Maybe it's 2/a+b?

@ohhwhat have you really never learned how to work with fractions? What level are you in school right now? What math courses are you taking at the same time as this introductory physics class where you got this optics problem?

 

[Mark44]

You may need to review how to add fractions. What is, for instance, $\dfrac{1}{a} + \dfrac{1}{b}$?

Maybe it's 2/a+b?

Life would be much simpler if it were only that easy.

I concur with @etotheipi that you should go back and review the arithmetic of fractions -- like immediately. You have absolutely zero hope of being able to finish this optics problem if you are unable to perform fraction arithmetic.

 

[ohhwhat]

@ohhwhat have you really never learned how to work with fractions? What level are you in school right now? What math courses are you taking at the same time as this introductory physics class where you got this optics problem?

Math courses where i don't need stuff like that anymore. I learned that a few years ago and cannot remember how it works. Also, I am especially very bad at maths, yikes. This "optics problem" is unfortunately part of my apprentice ship, although i have nothing to do with optics there.

 

[hutchphd]

So why are they asking this question? What is the apprenticeship?

 

[ohhwhat]

I suggest you write it as
\begin{align*}

\frac{1}{u} + \frac{1}{80-u} & = \frac{1}{f} \\ \\

\implies \frac{1}{u} \cdot \underbrace{\left(\frac{80-u}{80-u}\right)}_{=1} + \frac{1}{80-u} \cdot \underbrace{\left(\frac{u}{u}\right)}_{=1} &= \frac{1}{f} \\ \\

\implies \frac{80-u}{u(80-u)} + \frac{u}{u(80-u)} &= \frac{1}{f}

\end{align*}Do you see why? We re-wrote each fraction in terms of a common denominator, by sneakily multiplying each fraction by $1$ (which doesn't change the fraction).

Can you now simplify that last line, and then solve the equation for $u$?

So, the next line is: 80/u(80-u)=1/f , right?

 

[DaveE]

So, the next line is: 80/u(80-u)=1/f , right?

yes

 

[ohhwhat]

So why are they asking this question? What is the apprenticeship?

Idk what it's called in english, its "computer stuff", you know? I don't know why i need to have such a big knowledge about optics, nvm i have to do it anyways

 

[ohhwhat]

yes

Thanks! Then i finally understand the issues with fractions, i guess

 

[Orodruin]

Math courses where i don't need stuff like that anymore. I learned that a few years ago and cannot remember how it works. Also, I am especially very bad at maths, yikes. This "optics problem" is unfortunately part of my apprentice ship, although i have nothing to do with optics there.

Regardless of whether you will need to solve optics problems, you will have to solve equations similar to this and work with fractions to complete a degree in computer science. I strongly suggest you review basic math if you wish to stand a chance. Blindly guessing what the answer should be simply won't do.

 

[DaveE]

Thanks! Then i finally understand the issues with fractions, i guess

Have you studied algebra yet? Solving equations? What about quadratic equations, do you remember those?

 

[hutchphd]

The issue is to be able to do simple algebra! Now you will have to solve a quadratic equation for u (put in the value for f). This kind of problem will show up again and again.

 

[ohhwhat]

As solution for "u", i got u1=60, u2=20, is that correct?

 

[DaveE]

As solution for "u", i got u1=60, u2=20, is that correct?

yes, but why do you have to ask? Can't you just put those numbers into the original equation and check them yourself? The ability to check your own answers for correctness is an important skill in any STEM field. It's an essential habit to develop.

 

[ohhwhat]

I got it. You can close this.
Thanks to everyone who helped me!