$A$ convex lens of focal length $f$ is cut along the dashed line as shown in the figure. What is the focal length of each part?

$A$ convex lens is in contact with a concave lens. The magnitude of the ratio of their focal lengths is $2/3$. Their equivalent focal length is $30 \, cm$. Their individual focal lengths in $cm$ will be

Two thin equiconvex lenses each of focal length $0.2 \,m$ are placed coaxially with their optic centres $0.5 \,m$ apart. Then the focal length of the combination is (in $\,m$)

$A$ 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 . . . . . . .

Concave and convex lenses are placed touching each other. The ratio of the magnitudes of their power is $2:3$. The focal length of the system is $30 \ cm$. Then the focal lengths of the individual lenses are:

$A$ thin convex lens of focal length $30 \, cm$ forms an image $2 \, cm$ high,of an object at infinity. $A$ thin concave lens of focal length $20 \, cm$ is placed $26 \, cm$ from the convex lens on the side of the image. The height of the final image is.....$cm$.