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 (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

$(a)$ Determine the 'effective focal length' of the combination of two lenses,a convex lens of focal length $30 \; cm$ and a concave lens of focal length $20 \; cm$,if they are placed $8.0 \; cm$ apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?
$(b)$ An object $1.5 \; cm$ in size is placed on the side of the convex lens in the arrangement $(a)$ above. The distance between the object and the convex lens is $40 \; cm$. Determine the magnification produced by the two-lens system,and the size of the image.

$A$ convex lens of focal length $20 \, cm$ is cut into two equal parts. This results in two plano-convex lenses as shown in the figure. These two parts are then placed in contact with each other as shown in the figure. What will be the focal length of the system in $cm$?

An optician makes spectacles having a combination of a convex lens of focal length $40 \ cm$ in contact with a concave lens of focal length $25 \ cm$. The power of this combination of lenses in dioptre is:

Two plano-convex lenses,each of focal length $10 \, cm$ and refractive index $3/2$,are placed as shown in the figure. The space between them is filled with water $\left( \mu = 4/3 \right)$. The whole arrangement is in air. The optical power of the system is ....... $D$.