Galileo is credited with inventing the first telescope in 1610 - Edexcel - A-Level Physics - Question 18 - 2023 - Paper 2
Question 18
Galileo is credited with inventing the first telescope in 1610. The picture shows an early demonstration of the telescope.
A converging lens was positioned at one e... show full transcript
Worked Solution & Example Answer:Galileo is credited with inventing the first telescope in 1610 - Edexcel - A-Level Physics - Question 18 - 2023 - Paper 2
Step 1
a) Explain what can be concluded about the object being viewed.
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Answer
In observing distant objects through a telescope, the rays must be parallel, indicating that the object is at a significant distance from the lens. Therefore, the image created will be clear and distinct, suggesting that the object itself is far away.
Step 2
b) State what is meant by virtual and upright.
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Virtual: A virtual image is one that cannot be projected on a screen as the rays do not converge at the image location; they only appear to diverge from a point behind the lens.
Upright: An upright image maintains the same orientation as the original object, meaning that it is not inverted.
Step 3
c) Draw the ray diagram for the diverging lens.
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To draw the ray diagram for the diverging lens, start with an incoming parallel ray toward the lens, which diverges after passing through it. Draw another ray passing through the center of the lens which will continue in a straight line. Extend the diverging rays backward to locate the virtual image.
Step 4
d) Calculate the focal lengths of each lens.
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Given the magnification (M) of the telescope is 10 and the total distance (D) between the lenses is 90 cm:
Using the formula:
10 = \frac{f_c}{f_d}$$
Let focal length of the diverging lens be $f_d$, thus $f_c = 10 * f_d$.
Since the total distance is given by:
$$D = f_d + f_c = 90 \
(f_d + 10 f_d = 90) \
11 f_d = 90 \
f_d = \frac{90}{11} \approx 8.18 ext{ cm} \
f_c = 10 * f_d \approx 81.82 ext{ cm}$$
Step 5
e) Calculate the mass of Jupiter.
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Answer
To find the mass of Jupiter (M), use Kepler's third law:
T2=GM4π2a3
Where:
T=171 hours converted to seconds (171×3600=614760s).
a=1.07×106 km=1.07×109 m.
Substituting the values:
M=GT24π2a3
Using G=6.674×10−11 N m2 kg−2, calculate:
M=(6.674×10−11)(614760)24π2(1.07×109)3≈1.9×1027 kg