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A telscope has an objective lens of focal length 200 cm and an eye piece with focal length 2 cm . If this telescope is used to see 50m tall building at a distance of 2km, what is the height of the image of the building formed by the objective lens.?
Most Upvoted Answer
A telscope has an objective lens of focal length 200 cm and an eye pie...
By convex lens formula:1/v - 1/u =1/f
v = f*u/u-f ; 200*200*10³/200-200*10³ (2km=200*10³cm)
v= 200*10³/999
in magnification, {v/u} = {h'/h}
so, 200*10³/999*200*10³
h'=5000/999=5.005
h' is nearly equal to 5cm.
Community Answer
A telscope has an objective lens of focal length 200 cm and an eye pie...
Theory:

According to the lens formula, the magnification produced by a simple telescope is given by:

Magnification (M) = - (focal length of objective lens)/(focal length of eyepiece)

The angular magnification (A) can be calculated using the formula:

A = M * (angle subtended by the image at the eye)/(angle subtended by the object at the eye)

The height of the image formed by the objective lens can be calculated using the formula:

Height of image = Magnification * Height of object

Given:

Focal length of objective lens (f1) = 200 cm
Focal length of eyepiece (f2) = 2 cm
Height of object (h) = 50 m
Distance of object (d) = 2 km = 2000 m

Solution:

Step 1: Calculate the magnification (M)

M = - (f1/f2) = - (200 cm / 2 cm) = -100

Step 2: Calculate the angular magnification (A)

To calculate the angular magnification, we need to find the angles subtended by the image and the object at the eye.

The angle subtended by the object at the eye can be calculated using the formula:

Angle subtended by object at the eye = h/d

Angle subtended by object at the eye = 50 m / 2000 m = 0.025 radians

The angle subtended by the image at the eye can be calculated using the formula:

Angle subtended by image at the eye = (Magnification) * (angle subtended by object at the eye)

Angle subtended by image at the eye = (-100) * (0.025 radians) = -2.5 radians (Note: The negative sign indicates an inverted image)

Therefore, the angular magnification is:

A = -2.5 radians / 0.025 radians = -100

Step 3: Calculate the height of the image formed by the objective lens

Height of image = Magnification * Height of object

Height of image = (-100) * (50 m) = -5000 m (Note: The negative sign indicates an inverted image)

Therefore, the height of the image formed by the objective lens is 5000 meters.

Conclusion:

The height of the image formed by the objective lens of the telescope is 5000 meters. The image is inverted due to the nature of the lens system used in the telescope.
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A telscope has an objective lens of focal length 200 cm and an eye piece with focal length 2 cm . If this telescope is used to see 50m tall building at a distance of 2km, what is the height of the image of the building formed by the objective lens.?
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A telscope has an objective lens of focal length 200 cm and an eye piece with focal length 2 cm . If this telescope is used to see 50m tall building at a distance of 2km, what is the height of the image of the building formed by the objective lens.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A telscope has an objective lens of focal length 200 cm and an eye piece with focal length 2 cm . If this telescope is used to see 50m tall building at a distance of 2km, what is the height of the image of the building formed by the objective lens.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A telscope has an objective lens of focal length 200 cm and an eye piece with focal length 2 cm . If this telescope is used to see 50m tall building at a distance of 2km, what is the height of the image of the building formed by the objective lens.?.
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