An object \(AB\) is placed \(15\text{ cm}\) on the left of a convex lens \(P\) of focal length \(10\text{ cm}\). Another convex lens \(Q\) is now placed \(15\text{ cm}\) right of lens \(P\). If the focal length of lens \(Q\) is \(15\text{ cm}\), the final image is _______.

For lens \(P\):

\(\dfrac{1}{v_1} - \dfrac{1}{u_1} = \dfrac{1}{f_1}\)

\(\dfrac{1}{v_1} - \dfrac{1}{-15} = \dfrac{1}{10}\)

\(\dfrac{1}{v_1} = \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{1}{30} \Rightarrow v_1 = 30\text{ cm}\)

Magnification \(m_1 = \dfrac{v_1}{u_1} = \dfrac{30}{-15} = -2\)

The image formed by lens \(P\) acts as a virtual object for lens \(Q\). Since the distance between the lenses is \(15\text{ cm}\), the object distance for lens \(Q\) is:

\(u_2 = 30 - 15 = +15\text{ cm}\)

For lens \(Q\):

\(\dfrac{1}{v_2} - \dfrac{1}{u_2} = \dfrac{1}{f_2}\)

\(\dfrac{1}{v_2} - \dfrac{1}{15} = \dfrac{1}{15}\)

\(\dfrac{1}{v_2} = \dfrac{2}{15} \Rightarrow v_2 = 7.5\text{ cm}\)

The positive sign indicates that the final image is real and formed \(7.5\text{ cm}\) to the right of lens \(Q\).

Magnification \(m_2 = \dfrac{v_2}{u_2} = \dfrac{7.5}{15} = \dfrac{1}{2}\)

Total magnification \(m = m_1 \times m_2 = -2 \times \dfrac{1}{2} = -1\)

Since \(|m| = 1\), the size of the final image is the same as that of the object \(AB\).

Answer: real, formed at \(7.5\text{ cm}\) right of lens \(Q\), with a size same as that of \(AB\)