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Well After successfully getting some concave and convex lens I started observing. At first I saw that if we keep concave lens near to your eye and just after x distance away if we keep convex lens we get one enlarged image but then I just thought to do the same thing in opposite way. I tried putting convex lens near to my eye and concave at same distance x but I was not able to see enlarged image. Why is that so ? Is there any limitation or any reason behind this ?

With concave lens at first and then convex lens

enter image description here

With convex lens at first and then concave lens

enter image description here

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Why would you expect the images to be the same?

The lens formula (object distance $u$, image distance $v$ and focal length $f$; Cartesian sign convention) is: $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

If you consider two lenses $1$ and $2$, the latter at a distance $d$ from the former, then the first lens forms the image at $v_1$: $$\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$$ $$\implies v_1 = \frac{u_1f_1}{u_1 + f_1}$$ And the image due to the second lens is formed at $v_2$: $$\frac{1}{v_2} - \frac{1}{v_1-d} = \frac{1}{f_2}$$ $$\implies v_2 = \frac{(v_1-d)f_2}{v_1-d+f_2}$$

Now, we substitute for $v_1$: $$v_2 = \frac{(u_1-d)f_1f_2 - du_1f_2}{u_1(f_1 + f_2) + f_1f_2 - du_1 -df_1}$$

Now, notice that only if $d = 0$ (corresponding to lenses in contact) is the above expression symmetric under $f_1 \leftrightarrow f_2$ i.e. an exchange of the (different) lenses. In all other cases (with $d \ne 0$), an exchange of the lenses does produce a different $v_2$ (the offending terms being the right most ones in both the numerator and denominator), and therefore, a different image (and the magnification will also be different). After all, you can't use a telescope both ways.

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