MHB Sustainable Gardening Tips for Beginners

[DaalChawal]

 

[I like Serena]

Hi DaalChawal,

First a diagram:
\begin{tikzpicture}[scale=2]
\def\u{2.5}
\def\v{3.75}
\def\x{1.5}
\def\f{1.5}
\coordinate[label=above: A] (A) at ({-(\u+2)},1.5);
\coordinate[label=below: D] (D) at ({-(\u+2)},-1.5);
\coordinate[label=above: B] (B) at (-\u,.5);
\coordinate[label=below: C] (C) at (-\u,-.5);
\coordinate[label=F] (F) at (-\f,0);
\coordinate[label=F] (F') at (\f,0);

\coordinate[label=below: A'] (A') at ({\v-\x},{-\x/2});
\coordinate[label=above: D'] (D') at ({\v-\x},{\x/2});
\coordinate[label=below: B'] (B') at (\v,{-\x/2});
\coordinate[label=above: C'] (C') at (\v,{\x/2});

\draw (-6,0) -- (5,0);
\draw[ultra thick] (0,-2) -- (0,2) node[ above ] {+};
\filldraw (F) circle (.03);
\filldraw (F') circle (.03);
\draw (A) -- (B) -- (C) -- (D) -- cycle;
\draw (A') -- (B') -- (C') -- (D') -- cycle;
\draw[yshift=-0.1cm, latex-latex] ({-(\u+2)},0) -- node[ below ] {2} (-\u,0);
\draw[xshift=-0.1cm, latex-latex] ({-(\u+2)},1.5) -- node[ above left] {3} ({-(\u+2)},-1.5);
\draw[xshift=-0.1cm, latex-latex] (-\u,0.5) -- node[ above left] {1} (-\u,-0.5);
\draw[yshift=-0.1cm, latex-latex] ({\v-\x)},{\x/2}) -- node[ below ] {$x$} (\v,{\x/2});
\draw[xshift=0.1cm, latex-latex] (\v,{\x/2}) -- node[ above right ] {$x$} (\v,{-\x/2});
\draw[yshift=-0.1cm, latex-latex] (-\u,0) -- node[ below ] {$u$} (0,0);
\draw[yshift=-0.1cm, latex-latex] (0,0) -- node[ below ] {$v$} (\v,0);

\draw[help lines] (B) -- (B');
\draw[help lines] (B) -- (0,0.5) -- (B');
\draw[help lines] (B) -- (0,{-\f/(\u-\f)*0.5}) -- (B');
\draw[help lines] (A) -- (A');
\draw[help lines] (A) -- (0,1.5) -- (A');
\draw[help lines] (A) -- (0,{-\f/((\u+2)-\f)*1.5}) -- (A');

\end{tikzpicture}Let $x$ be the side length of the square in the image.
Let $f$ be the focal length of the lens.
Let $u$ be the distance of $BC$ to the lens and let $v$ be the distance of its image to the lens.
Then $u+2$ is the distance of $AD$ to the lens, and $v-x$ is the distance of its image to the lens.

The magnification of an object, which is the size of the image over the size of the object, is the same as the image distance over the object distance.
So the magnification of $BC$ is: $\frac{B'C'}{BC}=\frac x 1=\frac{v}{u}$.
The magnification of $AD$ is: $\frac{A'D'}{AD} = \frac x 3 = \frac{v-x}{u+2}$.
Furthermore, the lens formula tells us that $\frac 1u+\frac 1v=\frac 1f$ and $\frac 1{u+2}+\frac 1{v-x}=\frac 1f$.

So:
\begin{cases}\frac x 1=\frac{v}{u} \\ \frac x 3= \frac{v-x}{u+2} \\ \frac 1u+\frac 1v=\frac 1f \\ \frac 1{u+2}+\frac 1{v-x}=\frac 1f
\end{cases}

Solve?

 

[DaalChawal]

Thanks, I got it now.

 

[I like Serena]

I had actually made a couple of mistakes.
The image should be upside down .
And the image of $AD$ is closer instead of further away. That is, its image distance is $v-x$ instead of $v+x$.
It becomes apparent when we actually try to solve the equations.

I've updated the drawing and the formulas in my previous post.
And I've also added some extra help lines to show that it actually works and is to scale.

 

[DaalChawal]

Well sir, I only read your solution that what was the concept used and then I tried on my own and I got the answer