When a biconvex lens of glass having refractive index $1.47$ is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have a refractive index

When an object of height $12 \ cm$ is placed at a distance from a convex lens, an image of height $18 \ cm$ is formed on a screen. Without changing the positions of the object and the screen, if the lens is moved towards the screen, another clear image is formed on the screen. The height of this image is (in $cm$)

$A$ screen is placed $90 \ cm$ from an object. The image is formed on the screen by a convex lens placed at two different locations separated by $20 \ cm$. The focal length of the lens is approximately equal to: (in $cm$)

If a lens is moved towards the object from a distance of $40 \,cm$ to $30 \,cm$,the magnification of the image remains the same (numerically). The focal length of the lens is ......... $cm$.

$A$ concave lens of glass,refractive index $1.5$,has both surfaces of the same radius of curvature $R$. On immersion in a medium of refractive index $1.75$,it will behave as a:

In order to obtain a real image of magnification $2$ using a converging lens of focal length $20 \ cm,$ where should an object be placed in $cm$?