The focal length of a lens with a refractive index of $1.5$ in air is $0.30 \ m$. When it is immersed in a liquid of refractive index $\frac{4}{3}$,its focal length in the liquid will be ...... $cm$.

$f_V$ and $f_R$ are the focal lengths of a convex lens for violet and red light respectively,and $F_V$ and $F_R$ are the focal lengths of a concave lens for violet and red light respectively. Then:

$A$ turnip sits before a thin converging lens,outside the focal point of the lens. The lens is filled with a transparent gel so that it is flexible; by squeezing its ends toward its center [as indicated in figure $(a)$],you can change the curvature of its front and rear sides. What happens to the lateral height of the image?

$A$ biconvex lens of focal length $f$ and radii of curvature of both surfaces $R$ is made of a material of refractive index $n_{1}$. This lens is placed in a liquid of refractive index $n_{2}$. How will this lens behave?

If the image formed by a thin convex lens of power $P$ has magnification $m$,then the image distance $v$ is:

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.......$cm$.