What is the relative decrease in focal length of a lens for an increase in optical power by $0.1 \ D$ from $2.5 \ D$? [$D$ stands for dioptre]

$A$ thin equiconvex glass lens of refractive index $1.5$ has a power of $5 \,D$. When the lens is immersed in a liquid of refractive index $\mu$, it acts as a divergent lens of focal length $100 \,cm$. The value of $\mu$ of the liquid is:

$A$ diminished image of an object is to be obtained on a screen $1.0 \ m$ from it. This can be achieved by appropriately placing

$A$ thin convex lens is placed just above an empty vessel of depth $80 \,cm$. The image of a coin kept at the bottom of the vessel is thus formed $20 \,cm$ above the lens. If now water is poured in the vessel up to a height of $64 \,cm$, what will be the approximate new position of the image? Assume that the refractive index of water is $4/3$.

$A$ screen is placed $90 \ cm$ from an object. The image is formed on the screen by a convex lens placed at two different locations separated by $20 \ cm$. The focal length of the lens is approximately equal to: (in $cm$)

In an optics experiment,with the position of the object fixed,a student varies the position of a convex lens and for each position,the screen is adjusted to get a clear image of the object. $A$ graph between the object distance $|u|$ and the image distance $|v|$,from the lens,is plotted using the same scale for the two axes. $A$ straight line passing through the origin and making an angle of $45^{\circ}$ with the $x$-axis meets the experimental curve at $P$. The coordinates of $P$ will be