$A$ 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......$cm^2$

$A$ 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:

$A$ lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I.$ If an aperture of diameter $\frac{d}{2}$ in the central region of the lens is covered by a black paper,the focal length of the lens and the intensity of the image will be,respectively:

$A$ plane wave passes through a convex lens. The geometrical shape of the wavefront that emerges is

Consider a thin lens placed between a source $(S)$ and an observer $(O)$ (See figure). Let the thickness of the lens vary as $w(b) = w_0 - \alpha b^2$, where $b$ is the vertical distance from the pole. $w_0$ and $\alpha$ are constants. Using Fermat's principle, i.e., the time of transit for a ray between the source and observer is an extremum, find the condition that all paraxial rays starting from the source will converge at a point $O$ on the axis. Find the focal length.
$(ii)$ $A$ gravitational lens may be assumed to have a varying width of the form $W = K_1 \log \left( \frac{K_2}{b} \right)$ (where $b_{\min} < b < b_{\max}$). Show that an observer will see an image of a point object as a ring about the center of the lens with an angular radius $\beta = \sqrt{\frac{(n - 1)K_1 u}{v(u + v)}}$.

An object is placed at a distance of $20 \ cm$ from a convex lens of focal length $10 \ cm$. The image is formed on the other side of the lens at a distance of . . . . . . $cm$.