When a plane wavefront is incident on a convex lens,the emerging wavefront becomes . . . . . . .

$A$ 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:

The refractive index of the material of a double convex lens is $1.5$ and its focal length is $5 \,cm$. If the radii of curvature are equal, the value of the radius of curvature (in $cm$) is

$A$ plano-convex lens is made of refractive index of $1.6$. The radius of curvature of the curved surface is $60 \,cm$. The focal length of the lens is (in $\,cm$)

The distance between a real object and its real image formed by a converging lens is $56 \,cm$. The focal length $f$ of the lens is ............

$A$ converging lens of focal length $20\, cm$ and diameter $5\, cm$ is cut along the line $AB$. The part of the lens shown shaded in the diagram is now used to form an image of a point $P$ placed $30\, cm$ away from it on the line $XY$,which is perpendicular to the plane of the lens. The image of $P$ will be formed: