The refractive index of a glass convex lens is $1.5$. The radius of curvature of each of the two surfaces of the lens is $40 \ cm$. The ratio of the power of the lens when immersed in a liquid of refractive index $1.25$ to that when placed in air is:

The power of a thin convex lens placed in air is $+4 \ D$. The refractive index of the material of the convex lens is $\frac{3}{2}$. If this convex lens is immersed in a liquid of refractive index $\frac{5}{3}$, then

$A$ converging beam of rays is incident on a diverging lens. Having passed through the lens,the rays intersect at a point $15\, cm$ from the lens on the opposite side. If the lens is removed,the point where the rays meet will move $5\, cm$ closer to the lens. The focal length of the lens is $......\, cm$.

The diameter of a plano-convex lens is $6 \, cm$ and its thickness at the centre is $3 \, mm$. If the speed of light in the material of the lens is $2 \times 10^8 \, m/s$,the focal length of the lens is.......$cm$.

$A$ 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 $4/3$.

Letters $A, B, C$ and $D$ are written on a cardboard as shown in the figure. The cardboard is kept at a suitable distance behind a transparent empty glass of cylindrical shape. If the glass is now filled with water,one sees an inverted image of the pattern on the cardboard when looking through the glass. Ignoring magnification effects,the image would appear as: