$A$ convex lens of refractive index $1.5$ and focal length $f = 18 \ cm$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha \times f$. The value of $\alpha$ is . . . . . . . (Refractive index of water $= 4/3$)

Refractive index of a glass convex lens is $1.5$. The radius of curvature of each of the two surfaces of the lens is $40 \text{ cm}$. The ratio of the power of the lens when immersed in a liquid of refractive index $1.25$ to that when placed in air is

Look at the ray diagram shown. What will be the focal length of the $1^{st}$ and the $2^{nd}$ lens,if the incident light ray passes without any net deviation?

An object placed at a distance of $16 \ cm$ from a convex lens produces an image of magnification $m$ $(m > 1)$. If the object is moved towards the lens by $8 \ cm$, then again an image of magnification $m$ is obtained. The numerical value of the focal length of the lens is (in $cm$)

The focal lengths for violet,green,and red light rays are ${f_V}$,${f_G}$,and ${f_R}$ respectively. Which of the following is the true relationship?

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