$A$ thin equiconvex glass lens of refractive index $1.5$ has a power of $5 \,D$. When the lens is immersed in a liquid of refractive index $\mu$, it acts as a divergent lens of focal length $100 \,cm$. The value of $\mu$ of the liquid is:

The focal length of a biconvex lens made of glass with equal radii of curvature is $f$. If the lens is dipped in water, what will be its new focal length? (Take the refractive index of glass as $3/2$ and water as $4/3$)

$A$ plano-convex lens is made of material having refractive index $1.5$. The radius of curvature of the curved surface is $60 \ cm$. The focal length of the lens is . . . . . . $cm$.

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

The figure shows a convex lens cut symmetrically into two equal halves and separated laterally by a distance $h$. $A$ point object placed at a distance $30\, cm$ from the lens halves forms two real images separated by a distance $d$. $A$ plot of $d$ versus $h$ is shown in the figure. The focal length of the lens is ......$cm$.

Will the focal length of a lens for red light be more,same,or less than that for blue light?