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Inessa [10]
3 years ago
10

Now consider the case in which the object is between the lens and the focal point. Trace the P ray (use the label P1P1P_1 for th

e segment of the P ray on the same side as the object and P2P2P_2 for the segment on the opposite side) and the M ray. Draw the vectors for the incident rays starting from the tip of the object. The location and orientation of the vectors will be graded.

Physics
1 answer:
luda_lava [24]3 years ago
5 0

Answer:

 1 / f = 1 / p + 1 / q

Explanation:

For this exercise we will use two methods, an analytical method which is to use the constructor equation

                 1 / f = 1 / p + 1 / q

where f is the focal length, positive for converging lenses and negative for diverging lenses, p and q are the distance to the object and the image respectively

                 m = h’/ h = - q / p

where m is the magnification, h ’and h are the height of the image and the object.

We will also use a graphical method, where three rays will be traded

1) A ray that passes through the center of the lens and should not

2) a ray that passes through the focal length and comes out parallel to the lens

3) A ray that is horizontal and comes out through the focal from the other side

for this second method see the attachment

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- \frac{mR^2 }{I  } \ vAnswer:

a)      w = - \frac{m r }{I} v  ,  b)   W = - ½ m_woman R² (1 + m_woman R / I²) v²

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final instant. After starting to move

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the moment is preserved

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          W = - ½ m_woman R² (1 + m_woman R / I²) v²

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