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A teacher sets up a stand carrying a convex lens of focal length 15 cm at 20.5 cm mark on the optical bench. She asks the students to suggest the position of the screen on the optical bench so that a distinct image of a distant tree is obtained on it. What should be the correct position of the screen as suggested by the students and why?

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Demiurgos
We get the clearest image if there is no magnification. When we have no magnification the image and real object have the same size.
If we look at the diagram that I  attached we can see that:
[tex]\frac{h_i}{h_0}=\frac{d_i}{d_0}[/tex]
Two triangles that I marked are similar and from this we get:
[tex]\frac{h_i}{h_0}=\frac{d_i-f}{f}[/tex]
The image and the object must have the same height so we get:
[tex]\frac{h_i}{h_0}=\frac{d_i-f}{f};h_i=h_0\\ 1=\frac{d_i-f}{f}\\ d_i=2f[/tex]
This tells how far the screen should be from the lens. 
The position of the screen on the optical bench is:
[tex]S=20.5cm+2f=20.5+2\cdot 15cm=50.5cm[/tex]
Ver imagen Demiurgos

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