The objective glass of a telescope is an achromat of focal length $90\,cm$. The magnitudes of dispersive powers of the two lenses are $0.024$ and $0.036$. The focal lengths of the two lenses are:

$A$ screen is placed $90 \ cm$ from an object. The image is formed on the screen by a convex lens placed at two different locations separated by $20 \ cm$. The focal length of the lens is approximately equal to: (in $cm$)

The height of the image formed by a converging lens on a screen is $8\,cm$. For the same position of the object and screen,another image of size $12.5\,cm$ is formed on the screen by shifting the lens. The height of the object is...

The focal length of a thin biconvex lens is $20 \ cm$. When an object is moved from a distance of $25 \ cm$ in front of it to $50 \ cm$,the magnification of its image changes from $m_{25}$ to $m_{50}$. The ratio $\frac{m_{25}}{m_{50}}$ 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. What happens to the lateral height of the image?

$A$ point object is placed on the axis of a thin convex lens of focal length $0.05 \ m$ at a distance of $0.2 \ m$ from the lens and its image is formed on the axis. If the object is now made to oscillate along the axis with a small amplitude of $A \ cm,$ then what is the amplitude of oscillation of the image? [You may assume $\frac{1}{1+x} \approx 1-x,$ where $x << 1$]