The converging or diverging ability of a lens or mirror is called

In the displacement method,the distance between the object and the screen is $70 \; cm$ and the focal length of the lens is $16 \; cm$. Find the separation between the magnified and diminished image positions of the lens. (in $; cm$)

$A$ thin convex lens is made of two materials with refractive indices $n_1$ and $n_2$,as shown in the figure. The radii of curvature of the left and right spherical surfaces are equal. $f$ is the focal length of the lens when $n_1 = n_2 = n$. The focal length is $f + \Delta f$ when $n_1 = n$ and $n_2 = n + \Delta n$. Assuming $\Delta n \ll (n - 1)$ and $1 < n < 2$,which of the following statement$(s)$ is/are correct?
$(1)$ The relation between $\frac{\Delta f}{f}$ and $\frac{\Delta n}{n}$ remains unchanged if both the convex surfaces are replaced by concave surfaces of the same radius of curvature.
$(2)$ $\left|\frac{\Delta f}{f}\right| < \left|\frac{\Delta n}{n}\right|$
$(3)$ For $n = 1.5, \Delta n = 10^{-3}$ and $f = 20 \text{ cm}$,the value of $|\Delta f|$ will be $0.04 \text{ cm}$.
$(4)$ If $\frac{\Delta n}{n} < 0$ then $\frac{\Delta f}{f} > 0$.

$A$ lateral object of height $h_o = 0.5\, cm$ is placed on the optical axis of a bi-convex lens of focal length $f = 80\, cm$,at an object distance $u = -60\, cm$. The image formed is:

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

$A$ converging lens forms an image of an object on a screen. The image is real and twice the size of the object. If the positions of the screen and the object are interchanged,leaving the lens in the original position,the new image size on the screen is