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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)30°
- b)15°
- c)60°
- d)1°
Correct answer is option 'C'. 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...
To find the angle formed by the image of the tower, we can use the formula for angular magnification:
\(M = -\frac{f_o}{f_e}\)
Where:
\(M\) is the angular magnification
\(f_o\) is the focal length of the objective lens
\(f_e\) is the focal length of the eyepiece
Substituting the given values into the formula:
\(M = -\frac{150}{5} = -30\)
Now, to find the angle formed by the image of the tower, we can use the formula for angular size:
\(\theta = \frac{h}{d}\)
Where:
\(\theta\) is the angle formed by the image
\(h\) is the height of the tower
\(d\) is the distance from the tower to the observer
Substituting the given values into the formula:
\(\theta = \frac{50}{1000} = 0.05\)
Since the formula for angular magnification is negative, the image formed by the telescope is inverted. Therefore, the angle formed by the image of the tower is -0.05 radians.
\(M = -\frac{f_o}{f_e}\)
Where:
\(M\) is the angular magnification
\(f_o\) is the focal length of the objective lens
\(f_e\) is the focal length of the eyepiece
Substituting the given values into the formula:
\(M = -\frac{150}{5} = -30\)
Now, to find the angle formed by the image of the tower, we can use the formula for angular size:
\(\theta = \frac{h}{d}\)
Where:
\(\theta\) is the angle formed by the image
\(h\) is the height of the tower
\(d\) is the distance from the tower to the observer
Substituting the given values into the formula:
\(\theta = \frac{50}{1000} = 0.05\)
Since the formula for angular magnification is negative, the image formed by the telescope is inverted. Therefore, the angle formed by the image of the tower is -0.05 radians.
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Community Answer
A telescope has an objective lens of focal length 150 cm and an eyepie...
The tower PQ in figure (19.5) subtends an angle α on the objective. As uo is very large, the first image P Q' is formed in the focal plane of the objective.
tanα = α = P′Q′/f∘ ..(i)
The final image P"Q" subtends an angle β on the eyepiece (and hence on the eye). We have from the triangle P Q'E:
tanβ = β = ′P′Q′/EP ...(ii)
The telescope is set for normal adjustment so that the final image is formed at infinity, the first image P'Q' must be in the focal plane of the eyepiece.Then EP=fe .
Thus, equation (ii) becomes
tanβ=β=f∘/fe ...(iii)
dividing equation (iii) by (i)
β/α=f∘/fe...(iv)
from equation (i) The α=P′Q′/f∘
P′Q′=mh1 ...(v) where m is magnification of objective lens m=uv
where v is position of first image P'Q' and u is position of tower v=f∘=150cm,u=1000m substituting value of m in equation (v) m
P′Q′= 0.15/1000 x 50m=.0075m
α= 0.0075/0.15 = 0.05
value of β can be obtained from equation (v)
tanβ=.05.15×α=30×.05=1.5rad
β=56.3 which is closer to 60°
Fig 19.5
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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)30°b)15°c)60°d)1°Correct answer is option 'C'. 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)30°b)15°c)60°d)1°Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET 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)30°b)15°c)60°d)1°Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 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)30°b)15°c)60°d)1°Correct answer is option 'C'. Can you explain this answer?.
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)30°b)15°c)60°d)1°Correct answer is option 'C'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET 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)30°b)15°c)60°d)1°Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 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)30°b)15°c)60°d)1°Correct answer is option 'C'. Can you explain this answer?.
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