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.
1 answer:
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:
the resulting temperature is 23.37 ⁰C
Explanation:
Given;
mass of the iron, m₁ = 80 g = 0.08 kg
mass of the water, m₂ = 200 g = 0.2 kg
mass of the iron vessel, m₃ = 50 g = 0.05 kg
initial temperature of the iron, t₁ = 100 ⁰C
initial temperature of the water, t₂ = 20 ⁰C
specific heat capacity of iron, c₁ = 462 J/kg⁰C
specific heat capacity of water, c₂ = 4,200 J/kg⁰C
let the temperature of the resulting mixture = T
Apply the principle of conservation of energy;
heat lost by the hot iron = heat gained by the water
Therefore, the resulting temperature is 23.37 ⁰C