An object is placed at a distance of $20 \ cm$ from a convex lens of focal length $10 \ cm$. The image is formed on the other side of the lens at a distance of . . . . . . $cm$.

The word $KVPY$ is written on a board and viewed through different lenses such that the board is at a distance beyond the focal length of the lens.
Ignoring magnification effects,consider the following statements.
$(I)$ First image has been viewed from the planar side of a plano-concave lens and second image from the planar side of a plano-convex lens.
$(II)$ First image has been viewed from the concave side of a plano-concave lens and second image from the convex side of a plano-convex lens.
$(III)$ First image has been viewed from the concave side of a plano-concave lens and second image from the planar side of a plano-convex lens.
$(IV)$ First image has been viewed from the planar side of a plano-concave lens and second image from the convex side of a plano-convex lens.
Which of the above statements are correct?

In the displacement method,there are two positions of a lens for which we get a clear image on a screen. If the first position of the lens is at $40 \, cm$ from the object and the second position is at $80 \, cm$ from the object,the focal length of the lens is ......... $cm$.

In the figure,an air lens of radii of curvature $10 \ cm$ ($R_1 = 10 \ cm$,$R_2 = -10 \ cm$) is cut in a cylinder of glass $(\mu = 1.5)$. The focal length and the nature of the lens is:

Remote sensing satellites move in an orbit that is at an average height of about $500 \, km$ from the surface of the earth. The camera onboard one such satellite has a screen of area $A$ on which the images captured by it are formed. If the focal length of the camera lens is $50 \, cm$,then the terrestrial area that can be observed from the satellite is close to ............... $A$.

The graph between $\frac{1}{u}$ and $\frac{1}{v}$ for a thin convex lens in order to determine its focal length is plotted as shown in the figure. The refractive index of the lens is $1.5$ and both its surfaces have the same radius of curvature $R$. The value of $R$ will be in $cm$.
(Where $u =$ object distance,$v =$ image distance)