$A$ point source of light is placed on the principal axis between $F$ and $2F$ of a concave lens. On the other side,very far away,a screen is placed perpendicular to the principal axis. As the screen is brought closer towards the lens,

Calculate the focal length of the given lens in $cm$,if the magnification is $-0.5$.

$A$ double convex lens has two surfaces of equal radii $15 \, cm$ and refractive index $\mu = 1.5$. Its focal length is equal to ..... $cm$.

$A$ 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:

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

$A$ thin lens made of glass of refractive index $\mu_g = 1.5$ has a focal length of $12 \ cm$ in air. If it is immersed in water of refractive index $\mu_w = 4/3$,what will be its new focal length in $cm$?