In the displacement method,the distance between the object and the screen is $70 \, cm$. The focal length of the lens is $16 \, cm$. The distance between the positions of the lens for the magnified and diminished images will be ..... $cm$.

For a thin convex lens, the radii of curvature are $15 \,cm$ and $30 \,cm$ respectively. The focal length of the lens is $20 \,cm$. The refractive index of the material is:

$A$ convex lens of focal length $f$ is placed somewhere in between an object and a screen. The distance between the object and the screen is $x$. If the numerical value of the magnification produced by the lens is $m$, then the focal length of the lens is

$A$ convex lens of focal length $25 \ cm$ and made of glass with refractive index $1.5$ is immersed in water. The absolute change in focal length of the glass is [Use refractive index of water = $\frac{4}{3}$] (in $cm$)

$A$ lens forms real and virtual images of an object when the object is at $u_1$ and $u_2$ distances respectively. If the size of the virtual image is double that of the real image,then the focal length of the lens is (Take the magnification of the real image as $m$)

For an object placed at a distance $2.4\,m$ from a lens,a sharp focused image is observed on a screen placed at a distance $12\,cm$ from the lens. $A$ glass plate of refractive index $1.5$ and thickness $1\,cm$ is introduced between the lens and the screen such that the glass plate faces are parallel to the screen. By what distance should the object be shifted so that a sharp focused image is observed again on the screen (in $,m$)?