$A$ collimated beam of light of diameter $2 \ mm$ is propagating along the $x$-axis. The beam is required to be expanded into a collimated beam of diameter $14 \ mm$ using a system of two convex lenses. If the first lens has a focal length of $40 \ mm$, then the focal length of the second lens is . . . . . . $mm$.

We combined a convex lens of focal length $f_1$ and a concave lens of focal length $f_2$. Their combined focal length is $F$. The combination of these lenses will behave like a concave lens if:

$A$ convex lens and a concave lens, each with a focal length of $4 \,cm$, are separated by a distance of $6 \,cm$ along their axis. An object is placed $8 \,cm$ before the convex lens. The distance between the object and its final image is (in $\,cm$)

Two lenses of power $-15 D$ and $+5 D$ are in contact with each other. The focal length of the combination is

$A$ thin convex lens of focal length $5 \ cm$ and a thin concave lens of focal length $4 \ cm$ are combined together (without any gap) and this combination has magnification $m_1$, when an object is placed $10 \ cm$ before the convex lens. Keeping the positions of the convex lens and object undisturbed, a gap of $1 \ cm$ is introduced between the lenses by moving the concave lens away, which leads to a change in the magnification of the total lens system to $m_2$. The value of $\left|\frac{m_1}{m_2}\right|$ is . . . . . . .

For a combination of two convex lenses of focal lengths $f_1$ and $f_2$ to act as a glass slab, the distance of separation between them is