Determine the exaggeration produced by the two lens system, Physics

Q. (a) Determine the'effective  focal length' of the grouping of the two lenses in Exercise 9.10 if they are placed 8.0 cm a part with their principal axes  coincident. Performed the answer depend on which side of the combination a beam of parallel light is incident? Is the idea of effective focal length of this system useful at all? (b)T he object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance among the object and the convex lens is 40cm.  Determine the exaggeration produced by the two < lens system and the size of the image.

Answer

A Focal length of the convex lens, f1=30cm

A Focal length of the concave lens, f2=-20cm

The Distance between the two lenses d=8.0cm

(a)While the parallel beam of light is incident on the convex lens first:

As-per to the lens formula we have

2429_Determine the exaggeration produced by the two lens system.png

Where

U1=Object distance=∞

v1=Image distance

v2=Image distance

1528_Determine the exaggeration produced by the two lens system1.png

The image will perform as a real object for the convex lens.

Pertaining lens formula to the convex lens, we have:

1753_Determine the exaggeration produced by the two lens system2.png

Where,

U1=Object distance

=- (20+d)=-(20+8)=-28cm

V1=Image distance

2365_Determine the exaggeration produced by the two lens system3.png

Therefore the parallel incident beam appears to diverge from a point that is (420-4) 416 cm from the left of the centre of the combination of the two lenses.

The answer does depend on the side of the combination at which the parallel beam of light is incident. The notion of effectual focal length does not seem to be useful for this combination.

(b)Height of the image h 1=1.5cm

An Object distance from the side of the convex lens u1=-40cm

|u1|=40cm

As-per to the lens formula:

1447_Determine the exaggeration produced by the two lens system4.png

Where v1=Image distance

1017_Determine the exaggeration produced by the two lens system5.png

Magnification=120/40=3

Therefore the magnification due to the  convex lensis3.

The image produced by the convex lens acts as an object for the concave lens .According to the lens formula:

5_Determine the exaggeration produced by the two lens system6.png

Where

U2=Object distance

=+ (120-8)=112cm.

V2 =Image distance

1080_Determine the exaggeration produced by the two lens system7.png

Magnification 1335_Determine the exaggeration produced by the two lens system8.png

= 2240/92 x 1/112  = 20/92

Therefore the magnification due to the concave lens is 20/92.

The magnification produced by the grouping of the two lenses is calculated as

m x m'

= 3 x 20/92 = 60/92  = 0.652

The magnification of the grouping is given as

2113_Determine the exaggeration produced by the two lens system9.png

Where

h1= Object size=1.5cm

h2 = Size of the image

194_Determine the exaggeration produced by the two lens system10.png

Therefore the height of the image is 0.98 cm.

Posted Date: 6/27/2013 6:58:43 AM | Location : United States







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