4. A common axis passes through two thin lenses, A and B, which are separated by distance 3.0 cm. The focal length of A and B are 6.0 cm and 7.0 cm respectively. A beam of parallel ray is incident onto A. Determine the position of the image produced by B when the beam has passed through both the lenses.
If there are experimental image of this concept will help me out too. so plz mention also the site that shows the images...
If there are experimental image of this concept will help me out too. so plz mention also the site that shows the images...
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I always tackle problems of this sort by finding the image from the first lens.
As its focal length is 6 cm a parallel beam will form an image 6 cm beyond this lens.
Which also happens to be 4.0 cm to the right of the second lens.
As this is an image, it is identical in its properties to putting an object at the same point.
ie putting an object ( upside down of course) 4.0 cm to the right of a lens of focal length 7.0 cm
Now using 1/Do + 1/Di = 1/f
we get 1/4 + 1/Di = 1/7
1/Di = 1/7-1/4 = - 0.107
so Di = - 9.333cm
If you master the technique you can start to understand any system of multiple lenses and mirrors.
Each element produces an image either real or virtual, and this in turn is used as the object for the next element. The process continues until you have exhausted the chain of elements.
The negative sign shows that it is a virtual image It will be the same way up as the object and it will be on the same side as the object. That means that the total effect of both lenses together is to produce an upside down virtual image at a distance 9.333cm to the right of the second lens.
Unfortunately it isn't easy to send you a drawing to show how it all works. Sometimes I have drawn things and posted them elsewhere for people but I think you should try to understand this logic first.
Email me later if it still isn't clear.
As its focal length is 6 cm a parallel beam will form an image 6 cm beyond this lens.
Which also happens to be 4.0 cm to the right of the second lens.
As this is an image, it is identical in its properties to putting an object at the same point.
ie putting an object ( upside down of course) 4.0 cm to the right of a lens of focal length 7.0 cm
Now using 1/Do + 1/Di = 1/f
we get 1/4 + 1/Di = 1/7
1/Di = 1/7-1/4 = - 0.107
so Di = - 9.333cm
If you master the technique you can start to understand any system of multiple lenses and mirrors.
Each element produces an image either real or virtual, and this in turn is used as the object for the next element. The process continues until you have exhausted the chain of elements.
The negative sign shows that it is a virtual image It will be the same way up as the object and it will be on the same side as the object. That means that the total effect of both lenses together is to produce an upside down virtual image at a distance 9.333cm to the right of the second lens.
Unfortunately it isn't easy to send you a drawing to show how it all works. Sometimes I have drawn things and posted them elsewhere for people but I think you should try to understand this logic first.
Email me later if it still isn't clear.
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my email add is cocoatrix@yahoo,com
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