$A$ bi-concave symmetric lens made of glass has a refractive index of $1.5$. It has both surfaces with the same radius of curvature $R$. On immersion in a liquid of refractive index $1.75$,it will behave as a:

$A$ plano-convex lens is made of refractive index of $1.6$. The radius of curvature of the curved surface is $60 \,cm$. The focal length of the lens is (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 $\frac{\Delta f}{f} > 0$.

The refractive index of the material of a double convex lens is $1.5$ and its focal length is $5 \,cm$. If the radii of curvature are equal, the value of the radius of curvature (in $cm$) is

Derive the thin lens equation.

$A$ thin lens of material having refractive index $\mu = 1.5$ and focal length of $20 \ cm$ in air has two media of different refractive indices $\mu_1 = 1.2$ and $\mu_2 = 2.5$ covering the upper and lower halves of the lens,respectively,as shown in the figure. If an object is placed on the principal axis,then its two images will be formed,one after refraction from the upper part and the other after refraction from the lower part. Considering the object to be at $\infty$,the separation between the two images formed would be ...... $cm$.