## Question

The near point of a short-sighted person is 10 cm and he desires to read a book 30 cm away from him. The power of the lens to be used by him is:

### Solution

Here *u* = –30 cm, *v* = –10 cm and *f* = ?

Solving, we get; *f* = – 15 cm (concave lens)

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#### SIMILAR QUESTIONS

A luminous point is moving at speed *v _{0}* towards a spherical mirror, along its axis. Then the speed at which the image of this point object is moving is given by (with

*R*= radius of curvature and

*u*= object distance):

A slide projector gives a magnification of 10. If it projects a slide of dimensions 3 cm × 2 cm on a screen, the area of the image on the screen will be:

An equiconvex glass lens (a) has a focal length *f* and power *P*. It is cut into two symmetrical halves (b) by a plane containing the principal axis. The two pieces are recombined as shown in fig. (c) The power of new combination is:

Given There is an equiconvex lens with radius of each surface equal to 20 cm. There is air in the object space and water in the image space. The focal length of lens is:

A real image is formed by a convex lens. If we put it in contact with a concave lens and the combination again forms a real image, which of the following is true for the new image from the combination?

Focal length of a convex lens will be maximum for:

A concave mirror is focal length *f* in air is used in a medium of refractive index 2. What will be the focal length of the mirror in the medium?

A luminous object is placed at a distance of 30 cm from a convex lens of focal length 20 cm. On the other side of the lens, at what distance from the lens must a convex mirror of radius of curvature 10 cm be placed in order to have an upright image of the object coincident with it

The far-point of a short-sighted eye is 200 cm. The power of the lens is:

The distance between an object and the screen is 100 cm. A lens produces an image on the screen when placed at either of the positions 40 cm apart. The power of the lens is: