$A$ double convex lens made of glass has both radii of curvature of magnitude $20 \,cm$. Incident light rays parallel to the axis of the lens will converge to a point at a distance $L$ from the common pole $P$. The value of $L$ is [Refractive index of glass $= 1.5$].

$A$ 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 $4/3$.

An object has an image thrice of its original size when kept at $8 \ cm$ and $16 \ cm$ from a convex lens. The focal length of the lens is:

$A$ thin biconvex lens is made of glass $(\mu = 1.50)$ and both surfaces have a radius of curvature of $20 \ cm$. $A$ beam of incident light is parallel to the principal axis of the lens. The lens converges it at a distance $L \ cm$. Then $L = \dots$

Write the lens formula for a thin lens.

The image distance of an object placed $10 \ cm$ in front of a thin lens of focal length $+ 5 \ cm$ is......$cm$.