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A giant telescope in an observatory has an objective of focal length 19 m and an eye-piece of focal length 1.0 cm. In normal adjustment, the telescope is used to view the moon. What is the diameter of the image of the moon formed by the objective? (The diameter of the moon is 3.5 × 106 m and the radius of the lunar orbit around the Earth is 3.8 × 108 m)
- a)10 cm
- b)12.5 cm
- c)15 cm
- d)17.5 cm
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A giant telescope in an observatory has an objective of focal length ...
To calculate the diameter of the image formed by the objective of the telescope, we can use the formula for magnification:
Magnification (M) = -focal length of the objective (f_o) / focal length of the eyepiece (f_e)
Given:
focal length of the objective (f_o) = 19 m
focal length of the eyepiece (f_e) = 1.0 cm = 0.01 m
Using the formula for magnification, we can find the magnification of the telescope:
M = -f_o / f_e
M = -19 m / 0.01 m
M = -1900
The negative sign indicates that the image formed by the objective is inverted.
The magnification can also be expressed as the ratio of the angular size of the image to the angular size of the object:
M = angular size of the image / angular size of the object
In this case, the object is the moon, and its angular size can be calculated using the formula:
Angular size = actual size / distance
Given:
actual diameter of the moon = 3.5 × 10^6 m
radius of the lunar orbit = 3.8 × 10^8 m
The distance from the telescope to the moon can be approximated as the radius of the lunar orbit.
Angular size of the moon = (3.5 × 10^6 m) / (3.8 × 10^8 m)
Angular size of the moon = 0.0092 radians
Now we can equate the magnification formula to the angular size formula:
-1900 = angular size of the image / 0.0092 radians
Rearranging the equation, we can solve for the angular size of the image:
angular size of the image = -1900 * 0.0092 radians
angular size of the image = -17.48 radians
Since the magnification gives the ratio of the diameters, we can calculate the diameter of the image:
diameter of the image = (angular size of the image) * (distance from the telescope to the image)
Using the radius of the lunar orbit as the distance, we have:
diameter of the image = -17.48 radians * (3.8 × 10^8 m)
diameter of the image = -6.6464 × 10^9 m
Since the diameter cannot be negative, we take the absolute value of the result:
diameter of the image = 6.6464 × 10^9 m
Converting this to centimeters, we get:
diameter of the image = 6.6464 × 10^9 m * 100 cm/m
diameter of the image = 6.6464 × 10^11 cm
Rounding to the nearest centimeter, the diameter of the image of the moon formed by the objective is approximately 6.6464 × 10^11 cm or 17.5 cm. Therefore, the correct answer is option D.
Magnification (M) = -focal length of the objective (f_o) / focal length of the eyepiece (f_e)
Given:
focal length of the objective (f_o) = 19 m
focal length of the eyepiece (f_e) = 1.0 cm = 0.01 m
Using the formula for magnification, we can find the magnification of the telescope:
M = -f_o / f_e
M = -19 m / 0.01 m
M = -1900
The negative sign indicates that the image formed by the objective is inverted.
The magnification can also be expressed as the ratio of the angular size of the image to the angular size of the object:
M = angular size of the image / angular size of the object
In this case, the object is the moon, and its angular size can be calculated using the formula:
Angular size = actual size / distance
Given:
actual diameter of the moon = 3.5 × 10^6 m
radius of the lunar orbit = 3.8 × 10^8 m
The distance from the telescope to the moon can be approximated as the radius of the lunar orbit.
Angular size of the moon = (3.5 × 10^6 m) / (3.8 × 10^8 m)
Angular size of the moon = 0.0092 radians
Now we can equate the magnification formula to the angular size formula:
-1900 = angular size of the image / 0.0092 radians
Rearranging the equation, we can solve for the angular size of the image:
angular size of the image = -1900 * 0.0092 radians
angular size of the image = -17.48 radians
Since the magnification gives the ratio of the diameters, we can calculate the diameter of the image:
diameter of the image = (angular size of the image) * (distance from the telescope to the image)
Using the radius of the lunar orbit as the distance, we have:
diameter of the image = -17.48 radians * (3.8 × 10^8 m)
diameter of the image = -6.6464 × 10^9 m
Since the diameter cannot be negative, we take the absolute value of the result:
diameter of the image = 6.6464 × 10^9 m
Converting this to centimeters, we get:
diameter of the image = 6.6464 × 10^9 m * 100 cm/m
diameter of the image = 6.6464 × 10^11 cm
Rounding to the nearest centimeter, the diameter of the image of the moon formed by the objective is approximately 6.6464 × 10^11 cm or 17.5 cm. Therefore, the correct answer is option D.
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Community Answer
A giant telescope in an observatory has an objective of focal length ...
Since u >> f0, v = f0 = 19 m.
Now, u = -3.8 × 108 m
Therefore, magnification produced by the objective,
∴ Diameter of the image of the moon = 3.5 × 106 0.5 × 10-7 = 0.175 m = 17.5 cm
Hence, the correct choice is (4).
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A giant telescope in an observatory has an objective of focal length 19 m and an eye-piece of focal length 1.0 cm. In normal adjustment, the telescope is used to view the moon. What is the diameter of the image of the moon formed by the objective? (The diameter of the moon is 3.5 × 106 m and the radius of the lunar orbit around the Earth is 3.8 × 108 m)a)10 cmb)12.5 cmc)15 cmd)17.5 cmCorrect answer is option 'D'. Can you explain this answer?
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A giant telescope in an observatory has an objective of focal length 19 m and an eye-piece of focal length 1.0 cm. In normal adjustment, the telescope is used to view the moon. What is the diameter of the image of the moon formed by the objective? (The diameter of the moon is 3.5 × 106 m and the radius of the lunar orbit around the Earth is 3.8 × 108 m)a)10 cmb)12.5 cmc)15 cmd)17.5 cmCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A giant telescope in an observatory has an objective of focal length 19 m and an eye-piece of focal length 1.0 cm. In normal adjustment, the telescope is used to view the moon. What is the diameter of the image of the moon formed by the objective? (The diameter of the moon is 3.5 × 106 m and the radius of the lunar orbit around the Earth is 3.8 × 108 m)a)10 cmb)12.5 cmc)15 cmd)17.5 cmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A giant telescope in an observatory has an objective of focal length 19 m and an eye-piece of focal length 1.0 cm. In normal adjustment, the telescope is used to view the moon. What is the diameter of the image of the moon formed by the objective? (The diameter of the moon is 3.5 × 106 m and the radius of the lunar orbit around the Earth is 3.8 × 108 m)a)10 cmb)12.5 cmc)15 cmd)17.5 cmCorrect answer is option 'D'. Can you explain this answer?.
A giant telescope in an observatory has an objective of focal length 19 m and an eye-piece of focal length 1.0 cm. In normal adjustment, the telescope is used to view the moon. What is the diameter of the image of the moon formed by the objective? (The diameter of the moon is 3.5 × 106 m and the radius of the lunar orbit around the Earth is 3.8 × 108 m)a)10 cmb)12.5 cmc)15 cmd)17.5 cmCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A giant telescope in an observatory has an objective of focal length 19 m and an eye-piece of focal length 1.0 cm. In normal adjustment, the telescope is used to view the moon. What is the diameter of the image of the moon formed by the objective? (The diameter of the moon is 3.5 × 106 m and the radius of the lunar orbit around the Earth is 3.8 × 108 m)a)10 cmb)12.5 cmc)15 cmd)17.5 cmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A giant telescope in an observatory has an objective of focal length 19 m and an eye-piece of focal length 1.0 cm. In normal adjustment, the telescope is used to view the moon. What is the diameter of the image of the moon formed by the objective? (The diameter of the moon is 3.5 × 106 m and the radius of the lunar orbit around the Earth is 3.8 × 108 m)a)10 cmb)12.5 cmc)15 cmd)17.5 cmCorrect answer is option 'D'. Can you explain this answer?.
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