If the refractive index from air to glass is $\frac{3}{2}$ and that from air to water is $\frac{4}{3}$,then the ratio of focal lengths of a glass lens in water and in air is

$A$ rod of length $2 \text{ cm}$ makes an angle of $\frac{2 \pi}{3} \text{ rad}$ with the principal axis of a thin convex lens. The lens has a focal length of $10 \text{ cm}$ and is placed at a distance of $\frac{40}{3} \text{ cm}$ from the object as shown in the figure. The height of the image is $\frac{30 \sqrt{3}}{13} \text{ cm}$ and the angle made by it with respect to the principal axis is $\alpha \text{ rad}$. The value of $\alpha$ is $\frac{\pi}{n} \text{ rad}$,where $n$ is:

$A$ 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:

Explain the first focal point and second focal point for a lens.

In order to obtain a real image of magnification $2$ using a converging lens of focal length $20 \ cm,$ where should an object be placed in $cm$?

$A$ convex lens is immersed in water. How will its power $P$ change?