The radius of curvature of each surface of a convex lens having refractive index $1.8$ is $20 \ cm$. The lens is now immersed in a liquid of refractive index $1.5$. The ratio of power of the lens in air to its power in the liquid will be $x : 1$. The value of $x$ is $.....$

$A$ concave lens of focal length $f$ produces an image $(1/x)$ of the size of the object. The distance of the object from the lens is:

The height of the image formed by a converging lens on a screen is $8\,cm$. For the same position of the object and screen,another image of size $12.5\,cm$ is formed on the screen by shifting the lens. The height of the object is...

The focal length of a converging lens in air is $R$. If it is dipped in water of refractive index $1.33$,then its focal length will be around (Refractive index of lens material is $1.5$).

Consider the following statements regarding the real images formed with a converging lens.
$I$. Real images can be seen only if the image is projected onto the screen.
$II$. The real image can be seen only from the same side of the lens as that on which the object is positioned.
$III$. Real images produced by converging lenses are not only laterally but also longitudinally inverted as with mirrors.
Which of the above statement$(s)$ is/are incorrect?

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