The slit of a collimator is illuminated by a source as shown in the adjoining figures. The distance between the slit $S$ and the collimating lens $L$ is equal to the focal length of the lens. The correct direction of the emergent beam will be as shown in figure:

The magnitudes of power of a biconvex lens (refractive index $1.5$) and that of a plano-concave lens (refractive index $= 1.7$) are same. If the curvature of the plano-concave lens exactly matches with the curvature of the back surface of the biconvex lens, then the ratio of the radius of curvature of the front and back surface of the biconvex lens is . . . . . . .

$A$ thin double convex lens has radii of curvature each of magnitude $40 \; cm$ and is made of glass with refractive index $1.65$. Its focal length is nearly....... $cm$.

$A$ candle placed $25 \ cm$ from a lens forms an image on a screen placed $75 \ cm$ on the other side of the lens. The focal length and type of the lens should be

$A$ student measures the focal length of a convex lens by placing an object pin at a distance '$u$' from the lens and measuring the distance '$v$' of the image pin. The graph between '$u$' and '$v$' plotted by the student should look like:

$A$ 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 $4/3$.