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

Two thin lenses whose powers are $+2 \ D$ and $-4 \ D$ respectively are combined. The power of the combination is:

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:

$A$ 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 . . . . . . .

The focal length of a lens of dispersive power $0.45$ which should be placed in contact with a convex lens of focal length $84 \,cm$ and dispersive power $0.21$ to make the achromatic combination from the two lenses,in $cm$ is

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: