Two thin lenses of focal length $f_{1}$ and $f_{2}$ are in contact and coaxial. The power of the combination is . . . . . . .

Three lenses of focal lengths $+10 \ cm, -10 \ cm$ and $+30 \ cm$ are placed at distances of $30 \ cm, 35 \ cm$ and $45 \ cm$ respectively from an object. The distance between the object and the final image formed is: (in $cm$)

Two lenses of power $6D$ and $-2D$ are combined to form a single lens. The focal length of this lens will be (in $m$)

Two lenses $A$ and $B$ having focal lengths $2.0 \,cm$ and $5.0 \,cm$, respectively, are placed $14 \,cm$ apart. Lens $A$ is placed to the left of lens $B$. An object is placed $3 \,cm$ to the left of lens $A$. The distance of the final image from lens $A$ will be:

Two similar thin equi-convex lenses,each of focal length $f$,are kept coaxially in contact with each other such that the focal length of the combination is $F_{1}$. When the space between the two lenses is filled with glycerin (which has the same refractive index $\mu = 1.5$ as that of glass),the equivalent focal length is $F_{2}$. The ratio $F_{1} : F_{2}$ will be

The ray diagram for two lenses kept at some distance is given in the figure. Which of the following options is correct? ($f_1, f_2 =$ focal lengths,$d =$ distance between lenses)