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A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens.
Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.
Q. How far from the objective should an object be placed in order to obtain the condition described in part (i)?
- a)4.5 cm
- b)2.5 cm
- c)1.5 cm
- d)3.0 cm
Correct answer is option 'B'. Can you explain this answer?
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A compound microscope is an optical instrument used for observing hig...
Focal length of the objective lens, f1 = 2.0 cm
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Focal length of the eyepiece, f2 = 6.25 cm
Distance between the objective lens and the eyepiece, d = 15 cm
(a) Least distance of distinct vision, d' = 25
∴ Image distance for the eyepiece, v2 = −25 cm
Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
∴ u2 = −5 cm
Image distance for the objective lens, v1 = d + u2 = 15 − 5 = 10 cm
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
∴ u1 = −2.5 cm
Magnitude of the object distance, |u1| = 2.5 cm
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A compound microscope is an optical instrument used for observing hig...
Given:
Focal length of objective lens, f1 = 2.0 cm
Focal length of eyepiece, f2 = 6.25 cm
Distance between objective and eyepiece, d = 15 cm
To find:
Distance of object from the objective lens, u
Magnification produced by the objective lens, mo = -d/f1
Magnification produced by the eyepiece, me = 1 + d/f2
Total magnification, m = me x mo
We know that the magnifying power of a compound microscope is given by:
m = (θ' / θ) = (D / f1) x (1 + d / f2)
where,
θ' = angle subtended by the final image at the eye
θ = angle subtended by the object at the eye
D = least distance of distinct vision
For normal eye, D = 25 cm
Substituting the given values, we get:
m = (25 / 2) x (1 + 15 / 6.25)
m = 312.5
Also, m = me x mo
312.5 = me x (-15 / 2)
me = -41.67
As the magnification produced by the eyepiece is positive, we take the absolute value:
me = 41.67
Therefore, we can find the magnification produced by the objective lens as:
mo = m / me
mo = 7.5
Using the lens formula,
1/f = 1/v - 1/u
For the objective lens, the image is formed at the least distance of distinct vision, i.e., v = 25 cm
Substituting the given values, we get:
1/2 = 1/25 - 1/u
u = 2.5 cm
Hence, option B is the correct answer.
Focal length of objective lens, f1 = 2.0 cm
Focal length of eyepiece, f2 = 6.25 cm
Distance between objective and eyepiece, d = 15 cm
To find:
Distance of object from the objective lens, u
Magnification produced by the objective lens, mo = -d/f1
Magnification produced by the eyepiece, me = 1 + d/f2
Total magnification, m = me x mo
We know that the magnifying power of a compound microscope is given by:
m = (θ' / θ) = (D / f1) x (1 + d / f2)
where,
θ' = angle subtended by the final image at the eye
θ = angle subtended by the object at the eye
D = least distance of distinct vision
For normal eye, D = 25 cm
Substituting the given values, we get:
m = (25 / 2) x (1 + 15 / 6.25)
m = 312.5
Also, m = me x mo
312.5 = me x (-15 / 2)
me = -41.67
As the magnification produced by the eyepiece is positive, we take the absolute value:
me = 41.67
Therefore, we can find the magnification produced by the objective lens as:
mo = m / me
mo = 7.5
Using the lens formula,
1/f = 1/v - 1/u
For the objective lens, the image is formed at the least distance of distinct vision, i.e., v = 25 cm
Substituting the given values, we get:
1/2 = 1/25 - 1/u
u = 2.5 cm
Hence, option B is the correct answer.
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Community Answer
A compound microscope is an optical instrument used for observing hig...
Focal length of the objective lens, f1 = 2.0 cm
Focal length of the eyepiece, f2 = 6.25 cm
Distance between the objective lens and the eyepiece, d = 15 cm
(a) Least distance of distinct vision, d' = 25
∴ Image distance for the eyepiece, v2 = −25 cm
Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
∴ u2 = −5 cm
Image distance for the objective lens, v1 = d + u2 = 15 − 5 = 10 cm
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
∴ u1 = −2.5 cm
Magnitude of the object distance, |u1| = 2.5 cm
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A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens.Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.Q. How far from the objective should an object be placed in order to obtain the condition described in part (i)?a)4.5 cmb)2.5 cmc)1.5 cmd)3.0 cmCorrect answer is option 'B'. Can you explain this answer?
Question Description
A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens.Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.Q. How far from the objective should an object be placed in order to obtain the condition described in part (i)?a)4.5 cmb)2.5 cmc)1.5 cmd)3.0 cmCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens.Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.Q. How far from the objective should an object be placed in order to obtain the condition described in part (i)?a)4.5 cmb)2.5 cmc)1.5 cmd)3.0 cmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens.Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.Q. How far from the objective should an object be placed in order to obtain the condition described in part (i)?a)4.5 cmb)2.5 cmc)1.5 cmd)3.0 cmCorrect answer is option 'B'. Can you explain this answer?.
A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens.Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.Q. How far from the objective should an object be placed in order to obtain the condition described in part (i)?a)4.5 cmb)2.5 cmc)1.5 cmd)3.0 cmCorrect answer is option 'B'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens.Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.Q. How far from the objective should an object be placed in order to obtain the condition described in part (i)?a)4.5 cmb)2.5 cmc)1.5 cmd)3.0 cmCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens.Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.Q. How far from the objective should an object be placed in order to obtain the condition described in part (i)?a)4.5 cmb)2.5 cmc)1.5 cmd)3.0 cmCorrect answer is option 'B'. Can you explain this answer?.
Solutions for A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens.Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.Q. How far from the objective should an object be placed in order to obtain the condition described in part (i)?a)4.5 cmb)2.5 cmc)1.5 cmd)3.0 cmCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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ample number of questions to practice A compound microscope is an optical instrument used for observing highly magnified images of tiny objects. Magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the objects are situated at the least distance of distinct vision from the eye. It can be given that: m=me x mo, where me is the magnification produced by the eye lens and mo is the magnification produced by the objective lens.Consider a compound microscope that consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm.Q. How far from the objective should an object be placed in order to obtain the condition described in part (i)?a)4.5 cmb)2.5 cmc)1.5 cmd)3.0 cmCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice Class 12 tests.
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