$A$ thin lens is made of a material with a refractive index of $\mu = 1.5$. It is a biconvex lens. When it is immersed in water,it behaves as a/an:

$A$ 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:

Radii of curvature of a converging lens are in the ratio $1 : 2$. Its focal length is $6 \,cm$ and refractive index is $1.5$. Then its radii of curvature are ......... respectively.

An object placed $10 \ cm$ in front of a lens has an image $20 \ cm$ behind the lens. What is the power of the lens (in dioptres)?

$A$ convex lens of focal length $f$ produces a real image whose size is $n$ times the size of an object. The distance of the object from the lens is

$A$ convex lens of refractive index $1.5$ and focal length $f = 18 \ cm$ is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is $\alpha \times f$. The value of $\alpha$ is . . . . . . . (Refractive index of water $= 4/3$)