$A$ point object $O$ is placed on the principal axis of a convex lens of focal length $20 \, cm$ at a distance of $40 \, cm$ to the left of it. The diameter of the lens is $10 \, cm$. If the eye is placed $60 \, cm$ to the right of the lens at a distance $h$ below the principal axis,then the maximum value of $h$ to see the image will be.......$cm$.

An object approaches a convergent lens from the left of the lens with a uniform speed $5 \, m/s$ and stops at the focus. The image:

$A$ lens with a focal length of $50 \ cm$ forms an image of a distant object that subtends an angle of $1 \text{ milliradian}$ at the lens. What is the size of the image in $mm$?

An object is moving with a velocity $V_O = 4 \text{ cm/sec}$ towards a convex lens of focal length $f = 10 \text{ cm}$. The lens is also moving with a velocity $V_L = 2 \text{ cm/sec}$ in the same direction. At the instant when the object is at a distance of $30 \text{ cm}$ from the lens,find the velocity of the image.

$A$ 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:

$A$ 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. When you squeeze the lens,the image: