$A$ convex lens is dipped in a liquid whose refractive index is equal to the refractive index of the lens material. Then its focal length will

When the object is at distances $u_1$ and $u_2$ from a lens,the images formed are real and virtual respectively,and both have the same size. Then the focal length of the lens is:

The figure given below shows a beam of light converging at point $P.$ When a concave lens of focal length $16 \ cm$ is introduced in the path of the beam at a place $O$ shown by the dotted line such that $OP$ becomes the axis of the lens,the beam converges at a distance $x$ from the lens. The value of $x$ will be equal to.....$cm$.

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 . . . . . . .

$A$ convex lens of radii of curvature $6 \ cm$ and $12 \ cm$ is immersed in a liquid of refractive index $1.3$. If the refractive index of the material of the lens is $1.5$, then the focal length of the lens when immersed in the liquid is (in $cm$)

The diameter of a plano-convex lens is $6 \, cm$ and its thickness at the centre is $3 \, mm$. If the speed of light in the material of the lens is $2 \times 10^8 \, m/s$,the focal length of the lens is.......$cm$.