According to the question, the object is placed at 2F
The ray diagram is shown in the figure attached.
According to the figure:
Object AB is at 2F₁
First, we draw a ray parallel to principal axis.
So, it passes through focus after refraction.
We draw another ray which passes through optical center.
So, the ray will go through without any deviation.
Where both refracted rays meet is point A' and the image formed is A'B'
This image is formed at 2F₂
We can say that:
- Image is real.
- Image is inverted.
- Image is exactly the same size as the object.
Answer:
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<em>Hope this helps!!</em>
Answer:
the correct result is r = 3.71 10⁸ m
Explanation:
For this exercise we will use the law of universal gravitation
F = 
We call the masses of the Earth M, the masses of the moon m and the masses of the rocket m ', let's set a reference system in the center of the Earth, the distance from the Earth to the moon is d = 3.84 108 m
rocket force -Earth
F₁ = - \frac{m' M }{r^2}
rocket force - Moon
F₂ = - \frac{m' m }{(d-r)^2}
in the problem ask for what point the force has the relation
2 F₁ = F₂
let's substitute
2 
(d-r) ² = 
r²
d² - 2rd + r² = \frac{m}{2M} r²
r² (1 -\frac{m}{2M}) - 2rd + d² = 0
Let's solve this quadratic equation to find the distance r, let's call
a = 1 - \frac{m}{2M}
a = 1 - 
= 1 - 6.15 10⁻³
a = 0.99385
a r² - 2d r + d² = 0
r = 
r = [2d ± 2d 
] / 2a
r = 
(1 ± √ (1.65 10⁻³)) = (1 ± 0.04)
r₁ = \frac{d}{a} 1.04
r₂ = \frac{d}{a} 0.96
let's calculate
r₁ = 
1.04
r₁ = 401.8 10⁸ m
r₂ = \frac{3.84 10^8}{0.99385} 0.96
r₂ = 3.71 10⁸ m
therefore the correct result is r = 3.71 10⁸ m
Answer:
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Explanation:
Hope that helps :)
