$A$ concave lens of glass,refractive index $1.5$,has both surfaces of same radius of curvature $R$. On immersion in a medium of refractive index $1.75$,it will behave as a

The size of the images of an object,formed by a thin lens,are equal when the object is placed at two different positions $8 \ cm$ and $24 \ cm$ from the lens. The focal length of the lens is . . . . . . $cm$.

Parallel rays are incident on a thick plano-convex lens having radius of curvature $R$,refractive index $\mu$ and thickness $t$. When rays are incident on the plane surface,they converge at a distance $x$ from the plane surface. When rays are incident on the curved surface,they converge at a distance $y$ from the curved surface. Then:

The power of a biconvex lens is $1.25\,m^{-1}$ in a particular medium. The refractive index of the lens is $1.5$ and the radii of curvature are $20\,cm$ and $40\,cm$ respectively. Find the refractive index of the surrounding medium.

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 $\beta = \sqrt{\frac{(n - 1)K_1 u}{v(u + v)}}$.

$A$ lateral object of height $h_o = 0.5\, cm$ is placed on the optical axis of a bi-convex lens of focal length $f = 80\, cm$,at an object distance $u = -60\, cm$. The image formed is: