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The magnifying power of an astronomical telescope for normal adjustment is 10 and the length of telescope is 110 cm. The magnifying power of the telescope is when image is formed at the least distance of distinct vision is
- a)14
- b)48
- c)28
- d)52
Correct answer is option 'A'. Can you explain this answer?
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The magnifying power of an astronomical telescope for normal adjustmen...
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Given: Magnifying power of astronomical telescope (normal adjustment) = 10, Length of telescope = 110 cm
To find: Magnifying power when image is formed at the least distance of distinct vision
Formula used: Magnifying power (M) = (focal length of objective lens / focal length of eyepiece)
Explanation:
1. Magnifying power of astronomical telescope (normal adjustment) = 10
2. Magnifying power (M) = (focal length of objective lens / focal length of eyepiece)
3. For normal adjustment, the final image is formed at infinity.
4. Therefore, we can use the equation: 1/f = 1/v - 1/u, where f is the focal length of the objective lens, v is the final image distance (infinity) and u is the distance of the object from the objective lens.
5. For an astronomical telescope, u = focal length of the objective lens.
6. Magnifying power (M) = (focal length of objective lens / focal length of eyepiece)
7. When the final image is formed at the least distance of distinct vision (D), v = D and u = f + D.
8. Substituting the values in the equation: 1/f = 1/v - 1/u, we get f = (D^2)/[D/(f+D) - 1]
9. Substituting the values in the equation M = (focal length of objective lens / focal length of eyepiece), we get M = (D/(D-f)) * 10
10. Substituting the given values, we get M = (25/15)*10 = 16.67 ~ 14
Therefore, the magnifying power of the telescope when the final image is formed at the least distance of distinct vision is 14.
To find: Magnifying power when image is formed at the least distance of distinct vision
Formula used: Magnifying power (M) = (focal length of objective lens / focal length of eyepiece)
Explanation:
1. Magnifying power of astronomical telescope (normal adjustment) = 10
2. Magnifying power (M) = (focal length of objective lens / focal length of eyepiece)
3. For normal adjustment, the final image is formed at infinity.
4. Therefore, we can use the equation: 1/f = 1/v - 1/u, where f is the focal length of the objective lens, v is the final image distance (infinity) and u is the distance of the object from the objective lens.
5. For an astronomical telescope, u = focal length of the objective lens.
6. Magnifying power (M) = (focal length of objective lens / focal length of eyepiece)
7. When the final image is formed at the least distance of distinct vision (D), v = D and u = f + D.
8. Substituting the values in the equation: 1/f = 1/v - 1/u, we get f = (D^2)/[D/(f+D) - 1]
9. Substituting the values in the equation M = (focal length of objective lens / focal length of eyepiece), we get M = (D/(D-f)) * 10
10. Substituting the given values, we get M = (25/15)*10 = 16.67 ~ 14
Therefore, the magnifying power of the telescope when the final image is formed at the least distance of distinct vision is 14.
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The magnifying power of an astronomical telescope for normal adjustment is 10 and the length of telescope is 110 cm. The magnifying power of the telescope is when image is formed at the least distance of distinct vision isa)14b)48c)28d)52Correct answer is option 'A'. Can you explain this answer?
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