$A$ particle executes linear $S.H.M.$ along the principal axis of a convex lens of focal length $8 \,cm$. The mean position of oscillation is at $14 \,cm$ from the lens with amplitude $1 \,cm$. The amplitude of the oscillating image of the particle is nearly: (in $\,cm$)

Two point light sources are $24 \, cm$ apart. Where should a convex lens of focal length $9 \, cm$ be placed between them from one source so that the images of both the sources are formed at the same place (in $, cm$)?

$A$ 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 $1.5]$

$A$ convex lens is placed between an object and a screen which are at a fixed distance $D$ apart. For one position of the lens,the magnification of the image formed on the screen is $m_1$. When the lens is shifted by a distance $d$,the magnification of the image formed on the same screen is $m_2$. The focal length of the lens is:

The ratio of the focal lengths of a convex lens when kept in air and when it is immersed in a liquid is $1: 2$. If the refractive index of the material of the lens is $1.5$,then the refractive index of the liquid is

$A$ beam of light converges at a point $P$. Now a lens is placed in the path of the convergent beam $12 \, cm$ from $P$. At what point does the beam converge if the lens is
$(a)$ a convex lens of focal length $20 \, cm$,and
$(b)$ a concave lens of focal length $16 \, cm$?