$A$ point object is placed $20 \,cm$ to the left of a convex lens of focal length $f=5 \,cm$. The lens is made to oscillate with a small amplitude $A$ along the horizontal axis. The image of the object will also oscillate along the axis with:

$A$ 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 $\frac{\Delta f}{f} > 0$.

An equiconvex lens of radius of curvature $14 \ cm$ is made up of two different materials. The left half and right half of the vertical portion are made up of materials with refractive indices $1.5$ and $1.2$ respectively,as shown in the figure. If a point object is placed at a distance of $40 \ cm$,calculate the image distance. (in $cm$)

$A$ lens having refractive index $1.6$ has a focal length of $12 \ cm$ when it is in air. Find the focal length of the lens when it is placed in water. (Take the refractive index of water as $1.28$) (in $mm$)

$A$ screen is placed $90 \; cm$ from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by $20 \; cm$. Determine the focal length (in $cm$) of the lens.

The focal length of a thin biconvex lens is $20 \ cm$. When an object is moved from a distance of $25 \ cm$ in front of it to $50 \ cm$,the magnification of its image changes from $m_{25}$ to $m_{50}$. The ratio $\frac{m_{25}}{m_{50}}$ is