$A$ convex lens of focal length $40 \; cm$ is in contact with a concave lens of focal length $25 \; cm$. The power of the combination is:

One plano-convex and one plano-concave lens of same radius of curvature $R$ but of different materials are joined side by side as shown in the figure. If the refractive index of the material of $1$ is $\mu_1$ and that of $2$ is $\mu_2$,then the focal length of the combination is

Two lenses of power $+12 \ D$ and $-2 \ D$ are placed in contact. What will be the focal length of the combination in $cm$?

Curved surfaces of a plano-convex lens of refractive index $\mu_{1}$ and a plano-concave lens of refractive index $\mu_{2}$ have equal radius of curvature as shown in the figure. Find the ratio of the radius of curvature to the focal length of the combined lenses.

$A$ convex and a concave lens separated by a distance are then put in contact. How does the focal length of the combination change?

$A$ 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 . . . . . . .