A converging lens of focal length 6.25 cm is used as magnifying glass. If the near point of the observer is 25 cm from the eye and the lens is held close to the eye, calculate the distance of the object from the lens

A concave lens of focal length 15cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw ray diagram.

The near point of a hypermetropia eye is 1 m find the power of the lens required to correct this defect . Assume that near point of the normal eye is 25 cm

The size of the image of an object,which is at infinity, as formed by a convex lens of focal length 30 cm is 2 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance of 26 cm from the convex lens, calculate the new size of the image

If the distance between the screen and object is 50 cm and the displacement of the lens is 10 cm,clculate the focal length of the lens.

A card sheet divided into squares each of size 1 mm^2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to the eye. What should be the distance between the object in converging lens and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm^2 . Would you be able to see the squares distinctly with your eyes very close to the magnifier ?

If the focal length of a converging lens is 50 cm. What is the power of the Lens ?

A biconvex lens of focal length 15 cm is in front of a plane mirror.The distnace betweent the lens and the mirror is 10 cm.A small object is kept at a distance of 30 cm from the lens .The final image is