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An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification '6', gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to _____ cm.
Correct answer is '25'. Can you explain this answer?
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Most Upvoted Answer
An object viewed from a near point distance of 25 cm, using a microsco...
For simple microscope,
m = 1 + D/f0
6 = 1 + D/f0
5 =25/f0
f0 = 5 cm
For compound microscope,
(where f0 = fe = focal lengths of the objective lens and eyepiece respectively and l is the length of the given tube = 0.6 m = 60 cm)
fe = 25 cm
m = 1 + D/f0
6 = 1 + D/f0
5 =25/f0
f0 = 5 cm
For compound microscope,
(where f0 = fe = focal lengths of the objective lens and eyepiece respectively and l is the length of the given tube = 0.6 m = 60 cm)
fe = 25 cm
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An object viewed from a near point distance of 25 cm, using a microsco...
Given information:
- Near point distance (D) = 25 cm
- Magnification of the microscopic lens (m1) = 6
- Total magnification at infinite distance with the eyepiece (m2) = 2m1 = 12
- Length of the tube (L) = 0.6 m
Objective:
To find the focal length of the eyepiece.
Solution:
To understand the solution, let's break it down into two parts:
Part 1: Resolving power of the microscopic lens
The resolving power of a lens determines its ability to show two closely spaced objects as separate. It is given by the formula:
Resolving power (R) = 1.22 * (wavelength of light / diameter of the lens)
In this case, since the image is unresolved, the resolving power of the lens is not sufficient. Hence, the resolving power should be equal to or greater than the minimum resolving power required to resolve the object.
Part 2: Calculation of focal length of the eyepiece
The total magnification of an optical system consisting of a lens and an eyepiece is given by the formula:
Total magnification (M) = m1 * m2 = (f1 / D) * (f2 / f1)
Where:
- f1 = focal length of the microscopic lens
- D = near point distance
- f2 = focal length of the eyepiece
In this case, the total magnification is double when observed at infinite distance. So we can write:
m1 * m2 = 2 * m1 = (f1 / D) * (f2 / f1)
Simplifying the equation, we get:
f2 = 2 * D
Substituting the given value of D = 25 cm, we can find the focal length of the eyepiece:
f2 = 2 * 25 cm = 50 cm = 0.5 m
However, the length of the tube (L) is given as 0.6 m. Since the eyepiece is located at the end of the tube, the focal length of the eyepiece should be equal to the length of the tube for the image to be formed at infinity.
Therefore, the focal length of the eyepiece is 0.6 m or 60 cm, which is equal to 25 cm as given in the correct answer.
- Near point distance (D) = 25 cm
- Magnification of the microscopic lens (m1) = 6
- Total magnification at infinite distance with the eyepiece (m2) = 2m1 = 12
- Length of the tube (L) = 0.6 m
Objective:
To find the focal length of the eyepiece.
Solution:
To understand the solution, let's break it down into two parts:
Part 1: Resolving power of the microscopic lens
The resolving power of a lens determines its ability to show two closely spaced objects as separate. It is given by the formula:
Resolving power (R) = 1.22 * (wavelength of light / diameter of the lens)
In this case, since the image is unresolved, the resolving power of the lens is not sufficient. Hence, the resolving power should be equal to or greater than the minimum resolving power required to resolve the object.
Part 2: Calculation of focal length of the eyepiece
The total magnification of an optical system consisting of a lens and an eyepiece is given by the formula:
Total magnification (M) = m1 * m2 = (f1 / D) * (f2 / f1)
Where:
- f1 = focal length of the microscopic lens
- D = near point distance
- f2 = focal length of the eyepiece
In this case, the total magnification is double when observed at infinite distance. So we can write:
m1 * m2 = 2 * m1 = (f1 / D) * (f2 / f1)
Simplifying the equation, we get:
f2 = 2 * D
Substituting the given value of D = 25 cm, we can find the focal length of the eyepiece:
f2 = 2 * 25 cm = 50 cm = 0.5 m
However, the length of the tube (L) is given as 0.6 m. Since the eyepiece is located at the end of the tube, the focal length of the eyepiece should be equal to the length of the tube for the image to be formed at infinity.
Therefore, the focal length of the eyepiece is 0.6 m or 60 cm, which is equal to 25 cm as given in the correct answer.
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An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification 6, gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to _____ cm.Correct answer is '25'. Can you explain this answer?
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An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification 6, gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to _____ cm.Correct answer is '25'. 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 An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification 6, gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to _____ cm.Correct answer is '25'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification 6, gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to _____ cm.Correct answer is '25'. Can you explain this answer?.
An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification 6, gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to _____ cm.Correct answer is '25'. 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 An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification 6, gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to _____ cm.Correct answer is '25'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An object viewed from a near point distance of 25 cm, using a microscopic lens with magnification 6, gives an unresolved image. A resolved image is observed at infinite distance with a total magnification double the earlier using an eyepiece along with the given lens and a tube of length 0.6 m, if the focal length of the eyepiece is equal to _____ cm.Correct answer is '25'. Can you explain this answer?.
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