Q.Is it possible for two lenses to produce zero power?

The power of a lens is a measure of how strongly it converges or diverges light, defined as $P=1/f$ (SI unit dioptre, D; $1\,\text{D}=1\,\text{m}^{-1}$), positive for a converging lens and negative for a diverging one; from the lens maker's formula, $P=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$, so a higher refractive index or more sharply curved (smaller-radius, 'bulkier') surfaces give greater power (and hence a shorter focal length), while gently curved ('skinny') lenses have low power. For two thin lenses of focal lengths $f_1,f_2$ placed in contact (a common optical centre), writing and adding the lens equation for each in turn (the first lens's image serving as the second lens's object) gives the effective focal length $\dfrac{1}{F}=\dfrac{1}{f_1}+\dfrac{1}{f_2}$, extendable to any number of lenses as $\dfrac{1}{F}=\sum\dfrac1{f_i}$, or equivalently as an algebraic sum of powers $P=P_1+P_2+\ldots$ (some positive for convex, some negative for concave); the total magnification of the combination is the product of the individual magnifications, $m=m_1\times m_2\times\cdots$. Two thin lenses can even combine to give zero net power ($P_1=-P_2$) if a converging and a diverging lens of exactly matching magnitudes are used together. When two thin lenses of focal length $f_1,f_2$ are separated by a distance $d$ and only parallel incident rays (an object at infinity) are considered, tracing the net deviation angle through both lenses gives the special-case formula $\dfrac{1}{F}=\dfrac{1}{f_1}+\dfrac{1}{f_2}-\dfrac{d}{f_1f_2}$ for the combined focal length, together with expressions locating the position of the single equivalent lens measured from each of the two real lenses; this out-of-contact formula is valid only for the special case of an object at infinity -- for an object at a finite distance the image must instead be found by applying the lens equation separately, lens by lens, in sequence.