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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. The object distance for eye-piece, so that final image is formed at the least distance of distinct vision, will be
- a)3.45 cm
- b)5 cm
- c)1.29 cm
- d)2.59 cm
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
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
The magnifying power of a compound microscope is given by the relation:
= 4 × (1 + 4)
= 20
Hence, the magnifying power of the microscope is 20.
(b) The final image is formed at infinity.
∴ Image distance for the eyepiece, v2 = ∞
Object distance for the eyepiece = u2
According to the lens formula, we have the relation:
∴ u2 = −6.25 cm
Image distance for the objective lens, v1 = d + u2 = 15 − 6.25 = 8.75 cm
Object distance for the objective lens = u1
According to the lens formula, we have the relation:
Magnitude of the object distance, |u1| = 2.59 cm
Most Upvoted Answer
A compound microscope is an optical instrument used for observing hig...
To find the object distance for the eyepiece in a compound microscope, we need to use the formulas and principles of optics.
Given data:
Focal length of objective lens, fo = 2.0 cm
Focal length of eyepiece, fe = 6.25 cm
Separation between the lenses, d = 15 cm
To find the object distance for the eyepiece, we can use the lens formula:
1/f = 1/v - 1/u
where,
f = focal length of the lens
v = image distance
u = object distance
We can consider the objective lens as the first lens and the eyepiece as the second lens.
Let's assume the object distance for the eyepiece is u1 and the image distance formed by the objective lens is v1.
1) Calculation of image distance (v1) using the objective lens:
Using the lens formula for the objective lens:
1/fo = 1/v1 - 1/u1
Since the final image is formed at the least distance of distinct vision, the image distance v1 is equal to the least distance of distinct vision, which is generally taken as 25 cm.
Therefore, substituting the values:
1/2 = 1/25 - 1/u1
Simplifying the equation, we get:
1/u1 = 1/25 - 1/2
1/u1 = (2 - 25) / 50
1/u1 = -23 / 50
u1 = -50 / 23 cm
Note: The negative sign indicates that the image formed by the objective lens is virtual and on the same side as the object.
2) Calculation of object distance (u2) using the eyepiece:
Using the lens formula for the eyepiece:
1/fe = 1/v2 - 1/u2
Since the final image is formed at the least distance of distinct vision, the image distance v2 is also equal to the least distance of distinct vision, which is 25 cm.
Therefore, substituting the values:
1/6.25 = 1/25 - 1/u2
Simplifying the equation, we get:
1/u2 = 1/25 - 1/6.25
1/u2 = (6.25 - 25) / (6.25 * 25)
1/u2 = -18.75 / 156.25
u2 = -156.25 / 18.75 cm
u2 = -8.33 cm
Note: The negative sign indicates that the object for the eyepiece is placed on the same side as the final image.
3) Calculation of the object distance for the eyepiece (u):
Since the separation between the lenses is given as 15 cm, we can calculate the object distance for the eyepiece using the formula:
d = u2 - v1
Substituting the values:
15 = -8.33 - 25
15 = -33.33
This equation does not hold true.
There seems to be an error in the given data or calculation. Please recheck the values and calculations provided.
Given data:
Focal length of objective lens, fo = 2.0 cm
Focal length of eyepiece, fe = 6.25 cm
Separation between the lenses, d = 15 cm
To find the object distance for the eyepiece, we can use the lens formula:
1/f = 1/v - 1/u
where,
f = focal length of the lens
v = image distance
u = object distance
We can consider the objective lens as the first lens and the eyepiece as the second lens.
Let's assume the object distance for the eyepiece is u1 and the image distance formed by the objective lens is v1.
1) Calculation of image distance (v1) using the objective lens:
Using the lens formula for the objective lens:
1/fo = 1/v1 - 1/u1
Since the final image is formed at the least distance of distinct vision, the image distance v1 is equal to the least distance of distinct vision, which is generally taken as 25 cm.
Therefore, substituting the values:
1/2 = 1/25 - 1/u1
Simplifying the equation, we get:
1/u1 = 1/25 - 1/2
1/u1 = (2 - 25) / 50
1/u1 = -23 / 50
u1 = -50 / 23 cm
Note: The negative sign indicates that the image formed by the objective lens is virtual and on the same side as the object.
2) Calculation of object distance (u2) using the eyepiece:
Using the lens formula for the eyepiece:
1/fe = 1/v2 - 1/u2
Since the final image is formed at the least distance of distinct vision, the image distance v2 is also equal to the least distance of distinct vision, which is 25 cm.
Therefore, substituting the values:
1/6.25 = 1/25 - 1/u2
Simplifying the equation, we get:
1/u2 = 1/25 - 1/6.25
1/u2 = (6.25 - 25) / (6.25 * 25)
1/u2 = -18.75 / 156.25
u2 = -156.25 / 18.75 cm
u2 = -8.33 cm
Note: The negative sign indicates that the object for the eyepiece is placed on the same side as the final image.
3) Calculation of the object distance for the eyepiece (u):
Since the separation between the lenses is given as 15 cm, we can calculate the object distance for the eyepiece using the formula:
d = u2 - v1
Substituting the values:
15 = -8.33 - 25
15 = -33.33
This equation does not hold true.
There seems to be an error in the given data or calculation. Please recheck the values and calculations provided.
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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. The object distance for eye-piece, so that final image is formed at the least distance of distinct vision, will bea)3.45 cmb)5 cmc)1.29 cmd)2.59 cmCorrect answer is option 'D'. Can you explain this answer?
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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. The object distance for eye-piece, so that final image is formed at the least distance of distinct vision, will bea)3.45 cmb)5 cmc)1.29 cmd)2.59 cmCorrect answer is option 'D'. 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. The object distance for eye-piece, so that final image is formed at the least distance of distinct vision, will bea)3.45 cmb)5 cmc)1.29 cmd)2.59 cmCorrect answer is option 'D'. 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. The object distance for eye-piece, so that final image is formed at the least distance of distinct vision, will bea)3.45 cmb)5 cmc)1.29 cmd)2.59 cmCorrect answer is option 'D'. 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. The object distance for eye-piece, so that final image is formed at the least distance of distinct vision, will bea)3.45 cmb)5 cmc)1.29 cmd)2.59 cmCorrect answer is option 'D'. 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. The object distance for eye-piece, so that final image is formed at the least distance of distinct vision, will bea)3.45 cmb)5 cmc)1.29 cmd)2.59 cmCorrect answer is option 'D'. 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. The object distance for eye-piece, so that final image is formed at the least distance of distinct vision, will bea)3.45 cmb)5 cmc)1.29 cmd)2.59 cmCorrect answer is option 'D'. 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. The object distance for eye-piece, so that final image is formed at the least distance of distinct vision, will bea)3.45 cmb)5 cmc)1.29 cmd)2.59 cmCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 12.
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