When monochromatic red light is used instead of blue light in a convex lens,its focal length will .....

When an object is at a distance $u_1$ and $u_2$ from a lens,a real image and a virtual image are formed,respectively,having the same magnification $m$. The focal length of the lens is:

$A$ point source is located at a distance of $20 \ cm$ from the front surface of a symmetrical glass biconvex lens with equal radii of curvature $5 \ cm$. The distance at which the image is formed from the rear surface of this lens is $[$Given refractive index of the glass is $1.5]$

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

$A$ concave lens made of material of refractive index $1.6$ is immersed in a medium of refractive index $2.0$. The two surfaces of the concave lens have the same radius of curvature $0.2 \, m$. The lens will behave as a

In order to obtain a real image of magnification $2$ using a converging lens of focal length $20 \ cm,$ where should an object be placed in $cm$?