$A$ thin lens of focal length $f$ and its aperture has a diameter $d$. It forms an image of intensity $I$. Now, the central part of the aperture up to diameter $(d/2)$ is blocked by an opaque paper. The focal length and image intensity would change to:

An extended object is placed at point $O$,$10 \ cm$ in front of a convex lens $L_1$ and a concave lens $L_2$ is placed $10 \ cm$ behind it,as shown in the figure. The radii of curvature of all the curved surfaces in both the lenses are $20 \ cm$. The refractive index of both the lenses is $1.5$. The total magnification of this lens system is

$A$ biconvex lens has a radius of curvature of magnitude $20\, cm$. Which one of the following options best describes the image formed of an object of height $2\, cm$ placed $30\, cm$ from the lens?

The size of the images of an object, formed by a thin lens, are equal when the object is placed at two different positions $8 \ cm$ and $24 \ cm$ from the lens. The focal length of the lens is . . . . . . $cm$.

$A$ particle executes linear $S.H.M.$ along the principal axis of a convex lens of focal length $8 \,cm$. The mean position of oscillation is at $14 \,cm$ from the lens with amplitude $1 \,cm$. The amplitude of the oscillating image of the particle is nearly: (in $\,cm$)

If an object is placed at $A$ $(OA > f)$,where $f$ is the focal length of the lens,the image is found to be formed at $B$. $A$ perpendicular is erected at $O$ and $C$ is chosen on it such that the angle $\angle BCA$ is a right angle. Then the value of $f$ will be