Convex lens and concave lens are placed at a separation of 5cm each of f=10cm. If an object is placed at a distance of 30 cm in front of a convex lens, then find the position and nature of the final image formed.

Hint: Use the thin lens formula to find the image distance for the convex lens. Use this image distance as object distance for the concave lens to find the final position and nature of the image.

Formula used: $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$
Here
     v is image distance
     u is object distance
     f is focal length

Complete step by step solution:
For the convex lens
     $\begin{align}
  & {{u}_{1}}=-30\text{cm} \\
 & {{f}_{1}}=+10\text{cm( positive for convex lens)} \\
\end{align}$

Substituting in the formula
     $\begin{align}
  & \dfrac{1}{{{v}_{1}}}-\dfrac{1}{-30\text{cm}}=\dfrac{1}{10\text{cm}} \\
 & \dfrac{1}{{{v}_{1}}}=\dfrac{1}{10\text{cm}}-\dfrac{1}{\text{30cm}} \\
 & \dfrac{1}{{{v}_{1}}}=\dfrac{1}{15\text{cm}} \\
 & {{v}_{1}}=15\text{cm} \\
\end{align}$

The image is formed 15cm behind the convex lens. Concave lens is placed 5cm behind the convex lens. Therefore, the object for the concave lens is behind it.

For concave lens
     $\begin{align}
  & {{u}_{2}}=+10\text{cm} \\
 & {{f}_{2}}=-10\text{cm( negative for concave lens)} \\
\end{align}$





Substituting in the formula
     $\begin{align}
  & \dfrac{1}{{{v}_{2}}}-\dfrac{1}{10\text{cm}}=\dfrac{1}{-10\text{cm}} \\
 & \dfrac{1}{{{v}_{2}}}=\dfrac{1}{-10\text{cm}}+\dfrac{1}{\text{10cm}} \\
 & \dfrac{1}{{{v}_{2}}}=0 \\
 & {{v}_{2}}=\infty \\
\end{align}$
The image is formed at infinity.

A magnified image is formed at infinity for this system of convex and concave lens.

Note: The sign convention uses is the same as in cartesian coordinates, negative distance if it’s on the left of the origin (the convex or concave lens) and positive distance if it’s on the right. According to this convention, the focal length of a convex lens is positive and for a concave lens it is negative.