In the case of the displacement method of lenses,the product of magnification in both cases is

The magnitudes of power of a biconvex lens (refractive index $1.5$) and that of a plano-concave lens (refractive index $= 1.7$) are same. If the curvature of the plano-concave lens exactly matches with the curvature of the back surface of the biconvex lens, then the ratio of the radius of curvature of the front and back surface of the biconvex lens is . . . . . . .

$A$ boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length $10\, cm$. The diameter of the sun is $1.39 \times 10^9\, m$ and its mean distance from the earth is $1.5 \times 10^{11}\, m$. What is the diameter of the sun's image on the paper?

$A$ 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:

Two transparent media of refractive indices $\mu_1$ and $\mu_3$ have a solid lens-shaped transparent material of refractive index $\mu_2$ between them as shown in the figures in Column $II$. $A$ ray traversing these media is also shown in the figures. In Column $I$,different relationships between $\mu_1, \mu_2$,and $\mu_3$ are given. Match them to the ray diagram shown in Column $II$.

An object is placed at a distance of $5 \, cm$ from the first principal focus of a convex lens of focal length $10 \, cm$. If a real image is formed,find its distance from the lens in $cm$.