The dotted part of the lens is cut and kept on the $x$-axis as shown in the diagram. If parallel paraxial rays are falling on this system,then the coordinate of the image formed after refraction from both the lenses is $(30, -1)$. If $x = 2.5 \, cm$,then $y = .......... \, cm$. (Assume the lens has no spherical aberration)

The far point of a nearsighted person is $6.0 \, m$ from her eyes,and she wears contact lenses that enable her to see distant objects clearly. $A$ tree is $18.0 \, m$ away and $2.0 \, m$ high. How high is the image formed by the contact lenses?

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

The power of a thin convex lens placed in air is $+4 \ D$. The refractive index of the material of the convex lens is $\frac{3}{2}$. If this convex lens is immersed in a liquid of refractive index $\frac{5}{3}$, then

$A$ convex lens has a power of $+5.0 \, D$ in air,where the refractive index of the lens material is $_a\mu_g = 1.5$. In a liquid of what refractive index should it be immersed so that it acts as a concave lens of focal length $100 \, cm$?

The graph shows how the magnification $m$ produced by a thin lens varies with image distance $v$. What is the focal length of the lens used?