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:

Two thin biconvex lenses have focal lengths $f_{1}$ and $f_{2}$. $A$ third thin biconcave lens has a focal length of $f_{3}$. If the two biconvex lenses are in contact,the total power of the lenses is $P_{1}$. If the first convex lens is in contact with the third lens,the total power is $P_{2}$. If the second lens is in contact with the third lens,the total power is $P_{3}$,then:

$A$ convergent lens is placed $40 \,cm$ to the right of a diverging lens of focal length $15 \,cm$. $A$ parallel beam of light enters the diverging lens from the left, and the beam is again parallel when it emerges from the convergent lens. The focal length of the convergent lens is (in $\,cm$)

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

$A$ plano-convex lens (focal length $f_2,$ refractive index $\mu_2,$ radius of curvature $R$) fits exactly into a plano-concave lens (focal length $f_1,$ refractive index $\mu_1,$ radius of curvature $R$). Their plane surfaces are parallel to each other. Then,the focal length of the combination will be

$A$ 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 . . . . . . $mm$.