$A$ point object in air is in front of the curved surface of a plano-convex lens. The radius of curvature of the curved surface is $30 \; cm$ and the refractive index of the lens material is $1.5$. Then the focal length of the lens (in $cm$) is:

$A$ square card of side length $1\, mm$ is being seen through a magnifying lens of focal length $10\, cm$. The card is placed at a distance of $9\, cm$ from the lens. The axis is perpendicular to the plane of the card. The apparent area of the card through the lens is......$cm^2$

$A$ plano-concave lens is made of glass of refractive index $1.5$ and the radius of curvature of its curved face is $100 \,cm$. What is the power of the lens?

$A$ plano-convex lens having a radius of curvature of the first surface $2 \ cm$ exhibits a focal length of $f_1$ in air. Another plano-convex lens with a first surface radius of curvature $3 \ cm$ has a focal length of $f_2$ when it is immersed in a liquid of refractive index $1.2$. If both lenses are made of the same glass of refractive index $1.5$,the ratio of $f_1$ and $f_2$ will be:

$A$ thin convex lens made from crown glass $\left( \mu = \frac{3}{2} \right)$ has focal length $f$. When it is measured in two different liquids having refractive indices $\frac{4}{3}$ and $\frac{5}{3}$,it has the focal lengths $f_1$ and $f_2$ respectively. The correct relation between the focal lengths is:

The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall $3\;m$ away by means of a large convex lens. What is the maximum possible focal length (in $m$) of the lens required for the purpose?