Một kính lúp có ghi 5x trên vành của kính. Người quan sát có điểm cực cận cách mắt 20cm ngắm chừng ở vô cực để quan sát một vật. Số bội giác của kính là bao nhiêu? A. 4 B. 5 C. 2,5 D. 2
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{"content1": "The magnification of the lens can be calculated using the formula m = 1 + (d/f), where m is the magnification, d is the distance of the near point and f is the focal length of the lens.
In this case, the near point distance is 20cm. The focal length of the lens can be calculated using the formula f = 1/(focal power). Since the lens has a power of 5x, the focal length is 1/5 = 0.2m.
Substituting the values into the formula, we get m = 1 + (0.2/0.2) = 1 + 1 = 2.
Therefore, the magnification of the lens is 2.",
"content2": "The magnification of the lens can also be calculated using the formula m = d/(d-f), where m is the magnification, d is the distance of the near point and f is the focal length of the lens.
In this case, the near point distance is 20cm and the focal length of the lens is 0.2m. Substituting the values into the formula, we get m = 20/(20-0.2) ≈ 1.0025.
Therefore, the magnification of the lens is approximately 1.0025.",
"content3": "Another way to calculate the magnification of the lens is by using the formula m = d'/d, where m is the magnification, d' is the distance of the image formed by the lens and d is the distance of the object from the lens.
In this case, the distance of the image formed by the lens can be considered as infinity, so d' ≈ ∞. The distance of the object from the lens is 20cm. Substituting the values into the formula, we get m = ∞/20 ≈ ∞.
Therefore, the magnification of the lens can be considered as infinity.",
"content4": "The magnification of the lens can also be calculated using the formula m = (1 + d/f) / (1 + d/fe), where m is the magnification, d is the distance of the near point, f is the focal length of the lens and fe is the effective focal length of the lens.
Since the lens is a magnifying lens, the effective focal length can be calculated using the formula 1/fe = 1/f - 1/d. Substituting the values for f and d, we get 1/fe = 1/0.2 - 1/∞ ≈ 5 - 0 ≈ 5.
Substituting the values into the magnification formula, we get m = (1 + 0.2/0.2) / (1 + 0.2/5) ≈ 2.
Therefore, the magnification of the lens is approximately 2."}