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A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the angle formed by the image of the tower is θ, then θ is close to :
- a)1°
- b)15°
- c)30°
- d)60°
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
A telescope has an objective lens of focal length 150 cm and an eyepie...
We can use the formula for angular magnification of a telescope:
M = -(fo/fe)
where fo is the focal length of the objective lens and fe is the focal length of the eyepiece. The negative sign indicates that the image is inverted.
Plugging in the given values, we get:
M = -(150 cm)/(5 cm) = -30
This means that the image formed by the objective lens is magnified 30 times by the eyepiece.
To find the angle formed by the image, we can use the small angle approximation:
θ = h/d
where θ is the angle formed by the object, h is the height of the object, and d is the distance to the object.
Plugging in the given values, we get:
θ = (50 m)/(1000 m) = 0.05 radians
Now we can use the angular magnification to find the angle formed by the image:
θ' = θ/M
θ' = (0.05 radians)/(-30) = -0.0017 radians
The negative sign indicates that the image is inverted. To get the absolute value of the angle, we can take the magnitude:
|θ'| = 0.0017 radians
So the angle formed by the image of the tower is approximately 0.1 degrees.
M = -(fo/fe)
where fo is the focal length of the objective lens and fe is the focal length of the eyepiece. The negative sign indicates that the image is inverted.
Plugging in the given values, we get:
M = -(150 cm)/(5 cm) = -30
This means that the image formed by the objective lens is magnified 30 times by the eyepiece.
To find the angle formed by the image, we can use the small angle approximation:
θ = h/d
where θ is the angle formed by the object, h is the height of the object, and d is the distance to the object.
Plugging in the given values, we get:
θ = (50 m)/(1000 m) = 0.05 radians
Now we can use the angular magnification to find the angle formed by the image:
θ' = θ/M
θ' = (0.05 radians)/(-30) = -0.0017 radians
The negative sign indicates that the image is inverted. To get the absolute value of the angle, we can take the magnitude:
|θ'| = 0.0017 radians
So the angle formed by the image of the tower is approximately 0.1 degrees.
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A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the angle formed by the image of the tower is θ, then θ is close to :a)1°b)15°c)30°d)60°Correct answer is option 'D'. Can you explain this answer?
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A telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the angle formed by the image of the tower is θ, then θ is close to :a)1°b)15°c)30°d)60°Correct 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 telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the angle formed by the image of the tower is θ, then θ is close to :a)1°b)15°c)30°d)60°Correct 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 telescope has an objective lens of focal length 150 cm and an eyepiece of focal length 5 cm. If a 50 m tall tower at a distance of 1 km is observed through this telescope in normal setting, the angle formed by the image of the tower is θ, then θ is close to :a)1°b)15°c)30°d)60°Correct answer is option 'D'. Can you explain this answer?.
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