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If the focal length of objective and eye lens are 1.2 cm and 3 cm, respectively and the object is put 1.25 cm away from the objective lens, the final image is formed at infinity. The magnifying power of the microscope is
- a)400
- b)250
- c)200
- d)150
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
If the focal length of objective and eye lens are 1.2 cm and 3 cm, re...
The magnifying power of a microscope is given by the formula:
Magnifying power (M) = 1 + (D/F)
Where D is the least distance of distinct vision (25 cm) and F is the focal length of the objective lens.
In this case, the focal length of the objective lens (F1) is given as 1.2 cm and the focal length of the eye lens (F2) is given as 3 cm. The object is placed at a distance of 1.25 cm from the objective lens, and the final image is formed at infinity.
Let's calculate the magnifying power step by step:
Step 1: Calculate the distance of the object from the objective lens:
Distance of object (u) = 1.25 cm
Step 2: Calculate the distance of the image formed by the objective lens (v1):
Using the lens formula: 1/u + 1/v1 = 1/F1
Substituting the given values, we get: 1/1.25 + 1/v1 = 1/1.2
Solving the equation, we find: v1 = -6 cm
(Note: The negative sign indicates that the image formed by the objective lens is virtual and on the same side as the object)
Step 3: Calculate the distance of the image formed by the eye lens (v2):
Using the lens formula: 1/v1 + 1/v2 = 1/F2
Substituting the values, we get: 1/-6 + 1/v2 = 1/3
Solving the equation, we find: v2 = 6 cm
(Note: The positive sign indicates that the image formed by the eye lens is real and on the opposite side to the object)
Step 4: Calculate the magnifying power:
Magnifying power (M) = 1 + (D/F1)
Substituting the values, we get: M = 1 + (25/1.2) = 1 + 20.83 = 21.83
The magnifying power of the microscope is approximately 21.83, which is closest to option (C) 200.
Therefore, the correct answer is option (C) 200.
Magnifying power (M) = 1 + (D/F)
Where D is the least distance of distinct vision (25 cm) and F is the focal length of the objective lens.
In this case, the focal length of the objective lens (F1) is given as 1.2 cm and the focal length of the eye lens (F2) is given as 3 cm. The object is placed at a distance of 1.25 cm from the objective lens, and the final image is formed at infinity.
Let's calculate the magnifying power step by step:
Step 1: Calculate the distance of the object from the objective lens:
Distance of object (u) = 1.25 cm
Step 2: Calculate the distance of the image formed by the objective lens (v1):
Using the lens formula: 1/u + 1/v1 = 1/F1
Substituting the given values, we get: 1/1.25 + 1/v1 = 1/1.2
Solving the equation, we find: v1 = -6 cm
(Note: The negative sign indicates that the image formed by the objective lens is virtual and on the same side as the object)
Step 3: Calculate the distance of the image formed by the eye lens (v2):
Using the lens formula: 1/v1 + 1/v2 = 1/F2
Substituting the values, we get: 1/-6 + 1/v2 = 1/3
Solving the equation, we find: v2 = 6 cm
(Note: The positive sign indicates that the image formed by the eye lens is real and on the opposite side to the object)
Step 4: Calculate the magnifying power:
Magnifying power (M) = 1 + (D/F1)
Substituting the values, we get: M = 1 + (25/1.2) = 1 + 20.83 = 21.83
The magnifying power of the microscope is approximately 21.83, which is closest to option (C) 200.
Therefore, the correct answer is option (C) 200.
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If the focal length of objective and eye lens are 1.2 cm and 3 cm, re...
Magnifying power
= 
= 200
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If the focal length of objective and eye lens are 1.2 cm and 3 cm, respectively and the object is put 1.25 cm away from the objective lens, the final image is formed at infinity. The magnifying power of the microscope isa)400b)250c)200d)150Correct answer is option 'C'. Can you explain this answer?
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