$A$ lens placed $10 \,cm$ away from a wall casts a sharp inverted image of a candle on it. It again casts a sharp image when the lens is moved $20 \,cm$ further away from the wall. Now,the candle and the lens are moved such that a sharp inverted image with unit magnification is formed on the wall. To achieve this configuration,the candle was moved

$A$ convex lens of focal length $f$ produces a real image $n$ times the size of the object. The image distance is

$A$ glass lens is placed in a medium in which it is found to behave like a glass plate. The refractive index of the medium will be:

$A$ 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 focal length of a thin converging lens in air is $20 \ cm$. When the lens is immersed in a liquid,it behaves like a concave lens of power $1 \ D$. If the refractive index of the material of the lens is $1.5$,the refractive index of the liquid is

$A$ light ray hits the pole of a thin biconvex lens as shown in the figure. The angle made by the emergent ray with the optic axis will be