In the case of the displacement method of lenses,the product of magnification in both cases is

$A$ convex lens focuses an object $20 \ cm$ from it on a screen placed $5 \ cm$ away from it. $A$ glass plate (refractive index $\mu = \frac{7}{5}$) of thickness $t = 1.4 \ cm$ is inserted between the lens and the screen. What is the new distance of the object from the lens so that its image is again focused on the screen (in $cm$)?

$A$ convex lens of refractive index $1.5$ and focal length $18 \, cm$ in air is immersed in water. The change in focal length of the lens will be $........... \, cm$. (Given refractive index of water $= 4/3$)

$A$ real object approaches a fixed diverging lens with a constant velocity from infinity along the principal axis. The relative velocity between the object and its image will be

$A$ convex lens of refractive index $\frac{3}{2}$ has a power of $2.5 \ D$. If it is placed in a liquid of refractive index $2$, the new power of the lens is: (in $D$)

$A$ candle placed $25 \ cm$ from a lens forms an image on a screen placed $75 \ cm$ on the other side of the lens. The focal length and type of the lens should be