The radius of curvature of a convex lens is $40 \,cm$ for each surface. Its refractive index is $1.5$. Its focal length is: (in $\,cm$)

In the displacement method,the distance between the object and the screen is $70 \, cm$. The focal length of the lens is $16 \, cm$. The distance between the positions of the lens for the magnified and diminished images will be ..... $cm$.

The focal length of an equi-convex lens is greater than the radius of curvature of any of the surfaces. Then the refractive index of the material of the lens is

If the image formed by a thin convex lens of power $P$ has magnification $m$,then the image distance $v$ 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 $\beta = \sqrt{\frac{(n - 1)K_1 u}{v(u + v)}}$.

$A$ card sheet divided into squares each of size $1 \, mm^2$ is being viewed at a distance of $9 \, cm$ through a magnifying glass (a converging lens of focal length $10 \, cm$) held close to the eye. What is the magnification produced by the lens?