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Inessa [10]
4 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]4 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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Answer:

Speed:

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Wavelength:

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Explanation:

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Frenquency:

Since, wavelength is the only one who depends on the media. Therefore the frequency in both medium will be the same.  

To determine the frequency it can be used the following equation

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Where c is the speed of light, \nu is the frequency and \lambda is the wavelength

Then, \nu wil be isolated from equation 2.

\nu = \frac{c}{\lambda}  (3)

Before using equation 3 it is necessary to express \lamba in units of meters.

\lambda = 632.8nm . \frac{1m}{1x10^{9}nm} ⇒ 6.328x10^{-7}m

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\nu = 4.74x10^{14}Hz

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To determine the wavelength it can be used:

v = \nu \cdot \lambda

\lambda = \frac{v}{\nu}

Where v is the speed of the laser through the polystyrene.

\lambda = \frac{2.01x10^{8}m/s}{4.74x10^{14}s^{-1}}

\lambda = 4.24x10^{-7}m

Hence, the wavelength of the laser has a value of 4.24x10^{-7}m

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