The image of an illuminated square is obtained on a screen with the help of a converging lens. The distance of the square from the lens is $40\,cm$. The area of the image is $9$ times that of the square. The focal length of the lens is........$cm$.

$A$ thin converging lens of focal length $25 \,cm$ forms a sharp image of an object on a screen placed at a distance of $75 \,cm$ from the lens. Later, the screen is moved closer to the lens by a distance of $25 \,cm$. The distance through which the object is to be shifted so that its image on the screen is sharp again is

$A$ 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 $m$)

$A$ 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:

$A$ thin lens of refractive index $1.5$ has an optical power of $-5 \ D$ in air. Its optical power in a liquid medium with a refractive index of $1.6$ is $D$.

$A$ convex lens of glass $(\mu_g = 1.45)$ has a focal length $f_a$ in air. The lens is immersed in a liquid of refractive index $(\mu_l) = 1.3$. The ratio of the $f_l / f_a$ is