The plane faces of two identical plano-convex lenses,each having a focal length of $40 \ cm$,are pressed against each other to form a biconvex lens. The distance from this lens at which an object must be placed to obtain a real,inverted image with a magnification of $1$ is ....... $cm$.

$A$ plano-convex lens of refractive index $\mu_1$ fits exactly into a plano-concave lens of refractive index $\mu_2$. Their plane surfaces are parallel to each other. $R$ is the radius of curvature of the curved surface of the lenses. The focal length of the combination is:

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

Two thin lenses have a combined power of $+9 \text{ D}$. When they are separated by a distance of $20 \text{ cm}$,their equivalent power becomes $+\frac{27}{5} \text{ D}$. The power of both the lenses in dioptre are respectively:

An object is located $4 \ m$ from the first of two thin converging lenses of focal lengths $2 \ m$ and $1 \ m,$ respectively. The lenses are separated by $3 \ m$. The final image formed by the second lens is located from the source at a distance of (in $m$):

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