Diverging Lens

Diverging lens which are thinner in the middle than at the edge
Types of diverging lenses

The symbolize of diverging lenses

The part of diverging Lenses

Construction of rays diagram for diverging lenses

1.      A ray parallel to the principle axis is refracted seemed to the second focal point (F2)

2.      A ray proceeds toward the second focal point (F2) is  refracted parallel to the principle axis

3.      A ray passing trough to the center (C) is not deviated

The image formed by a concave lens not depends on the positions of the object

The image formed by a concave lens not depends on the positions of the object

Images of real objects formed by diverging lenses are always virtual, erect, and diminished in size

 

Images of real objects formed by diverging lenses are always virtual, erect, and diminished in size

The lens equations can be applied to either converging/convex or diverging/concave lenses

                 1/s + 1/s^’   =   1/f                    

                   Where:    s   = the distance of object

                                   s^’   =  the distance of  image

                                    f    = the distance of the focal point

 Magnification (M) is define as ratio between distance of the image(s^’) to distance of the object (s) or ratio between height of the image (h^’) to height of the object (h)

                In mathematically can be formulated as follows:

                 M = s^’/s    or M = h^’/h

The power of lens is ability of lens in collecting or spreading the ray it receives

 The power of lens is formulated:

                        P = 1/f

   In which:     P = power lens (diopters)

                       f = focal point (meter)

 Example 1

Question

 A diverging lens of the power of lens is 6  2/3 diopter   is placed 60 cm from the center of the lens. Locate and describe the image formed by the lens.

Solution:

Remember the power of lens is formulated:

Recalling that the power of lens is negative for diverging lens, we can substitute

                        P = 1/f

                         f = 1/-P

                           = 1/ -6 2/3

                           = 1/ -20/3

                           = 1 x -3/20 meter

                           = 1 x -300/20 cm

                           = -15 cm

f = - 15 cm and s = 60 cm into the lens equation

         1/s^’ = 1/f – 1/s

                = 1/-15 -1/60

                = - 4/60 – 1/60

                = - 5/60

            S^’ = -1/12

                = - 12 cm

Since the image distance is negative, we have confirmed the image is virtual. The nature is virtual, erect and diminished.

Example 2

Question

What are the nature and location of an image formed by the converging lens if an object is placed 15 cm from the center of the lens if the focal length is 10 cm?

What are magnification and height of the image, if the height of object is 2 cm?

Solution:

The object distance (s) = 15 cm, and the image distance can be found by solving the lenses equation explicitly for (s^’)

             1/s + 1/s^’   =   1/f  

             1/s^’ = 1/f – 1/s

                s^’  = s.f /s-f

Substituting f = + 10 cm and s = + 15 cm, we obtain

                s^’  = (15 cm ) ( 10 cm )/ 15 cm – 10 cm

                    = + 30 cm

The positive sign for s^’ verifies that the image is real. The nature is real, inverted and enlarged.

Magnification of the image can be found as follows,

                M = s^’/ s

                     = 30/10

                     = 3 times

The height of the image can be found as follows,

               h’ = M x h

                   = 3 x 2 cm

                   = 6 cm

EXRECISES

1. A converging lens has the power of lens is 5 diopter. If  an object which its height is 3 cm  is placed 30 cm in front of the lens, then determine :

a.       The focal length of the converging lens

b.      The distance of the image from the lens

c.       The magnification of the image

d.      The height of image

e.       Sketch rays diagram with scale

f.        The nature of the image

2. A diverging lens has the power of lens is 2,5  diopter. If  an object which its height is 3 cm  is placed 60 cm in front of the lens, then determine :

a.       The focal length of the converging lens

b.      The distance of the image from the lens

c.       The magnification of the image

d.      The height of image

e.       Sketch rays diagram with scale

f.        The nature of the image