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A collimated beam of light of diameter 1 mm is propagating along the x- axis. The beam is to be expanded 10 mm using a combination of two convex lenses. A lens of focal length of 50 mm and another lens with focal length F are to be kept at a distance d between them. The values of F and d respectively, are
- a)450 mm and 10 mm
- b)400 mm and 500 mm
- c)550 mm and 600 mm
- d)500 mm and 550 mm
Correct answer is option 'D'. Can you explain this answer?
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A collimated beam of light of diameter 1 mm is propagating along the x...
To expand the collimated beam of light, we can use a combination of two convex lenses. Let's calculate the values of the focal length (F) and the distance between the lenses (d) required for the desired expansion.
1. Given:
- Diameter of the collimated beam = 1 mm
- Expansion required = 10 mm
- Focal length of the first lens = 50 mm
2. Calculation of the focal length of the second lens:
The first lens would act as a diverging lens to spread out the beam, and the second lens would act as a converging lens to bring the beam back to its original diameter. Since the beam needs to be expanded, the second lens should have a focal length greater than 50 mm.
3. Calculation of the focal length of the second lens using lens formula:
The lens formula is given by 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance.
- For the first lens, the object distance (u) is the distance at which the light rays from the collimated beam converge to a point. Since the beam is collimated, the object distance would be infinity (u = ∞).
- The image distance (v) would be the distance at which the light rays from the first lens converge to a point on the second lens. This distance is equal to the distance between the lenses (d).
Using the lens formula, we can write 1/f1 = 1/d - 1/∞.
Since 1/∞ is zero, we have 1/f1 = 1/d, and rearranging the equation gives f1 = d.
- For the second lens, the object distance (u) would be the distance at which the light rays from the first lens converge to a point on the second lens. This distance is equal to the distance between the lenses (d).
- The image distance (v) is the distance at which the light rays from the second lens converge to form the expanded beam. This distance is equal to the expansion required (10 mm).
Using the lens formula again, we can write 1/f2 = 1/10 - 1/d.
4. Substituting the value of f1 = d into the equation for f2, we get:
1/f2 = 1/10 - 1/f1.
Substituting f1 = d, we have 1/f2 = 1/10 - 1/d.
5. Solving the equation for f2, we find:
1/f2 = (d - 10)/10d.
Taking the reciprocal of both sides, we get f2 = 10d/(d - 10).
6. As per the options given, we need to find the values of F and d that satisfy the equation f2 = 10d/(d - 10) = 500 mm and 550 mm, respectively. Plugging these values into the equation, we find that they satisfy the equation.
Therefore, the correct answer is option D: 500 mm and 550 mm.
1. Given:
- Diameter of the collimated beam = 1 mm
- Expansion required = 10 mm
- Focal length of the first lens = 50 mm
2. Calculation of the focal length of the second lens:
The first lens would act as a diverging lens to spread out the beam, and the second lens would act as a converging lens to bring the beam back to its original diameter. Since the beam needs to be expanded, the second lens should have a focal length greater than 50 mm.
3. Calculation of the focal length of the second lens using lens formula:
The lens formula is given by 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance.
- For the first lens, the object distance (u) is the distance at which the light rays from the collimated beam converge to a point. Since the beam is collimated, the object distance would be infinity (u = ∞).
- The image distance (v) would be the distance at which the light rays from the first lens converge to a point on the second lens. This distance is equal to the distance between the lenses (d).
Using the lens formula, we can write 1/f1 = 1/d - 1/∞.
Since 1/∞ is zero, we have 1/f1 = 1/d, and rearranging the equation gives f1 = d.
- For the second lens, the object distance (u) would be the distance at which the light rays from the first lens converge to a point on the second lens. This distance is equal to the distance between the lenses (d).
- The image distance (v) is the distance at which the light rays from the second lens converge to form the expanded beam. This distance is equal to the expansion required (10 mm).
Using the lens formula again, we can write 1/f2 = 1/10 - 1/d.
4. Substituting the value of f1 = d into the equation for f2, we get:
1/f2 = 1/10 - 1/f1.
Substituting f1 = d, we have 1/f2 = 1/10 - 1/d.
5. Solving the equation for f2, we find:
1/f2 = (d - 10)/10d.
Taking the reciprocal of both sides, we get f2 = 10d/(d - 10).
6. As per the options given, we need to find the values of F and d that satisfy the equation f2 = 10d/(d - 10) = 500 mm and 550 mm, respectively. Plugging these values into the equation, we find that they satisfy the equation.
Therefore, the correct answer is option D: 500 mm and 550 mm.
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A collimated beam of light of diameter 1 mm is propagating along the x- axis. The beam is to be expanded 10 mm using a combination of two convex lenses. A lens of focal length of 50 mm and another lens with focal length F are to be kept at a distance d between them. The values of F and d respectively, area)450 mm and 10 mmb)400 mm and 500 mmc)550 mm and 600 mmd)500 mm and 550 mmCorrect answer is option 'D'. Can you explain this answer?
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A collimated beam of light of diameter 1 mm is propagating along the x- axis. The beam is to be expanded 10 mm using a combination of two convex lenses. A lens of focal length of 50 mm and another lens with focal length F are to be kept at a distance d between them. The values of F and d respectively, area)450 mm and 10 mmb)400 mm and 500 mmc)550 mm and 600 mmd)500 mm and 550 mmCorrect answer is option 'D'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A collimated beam of light of diameter 1 mm is propagating along the x- axis. The beam is to be expanded 10 mm using a combination of two convex lenses. A lens of focal length of 50 mm and another lens with focal length F are to be kept at a distance d between them. The values of F and d respectively, area)450 mm and 10 mmb)400 mm and 500 mmc)550 mm and 600 mmd)500 mm and 550 mmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A collimated beam of light of diameter 1 mm is propagating along the x- axis. The beam is to be expanded 10 mm using a combination of two convex lenses. A lens of focal length of 50 mm and another lens with focal length F are to be kept at a distance d between them. The values of F and d respectively, area)450 mm and 10 mmb)400 mm and 500 mmc)550 mm and 600 mmd)500 mm and 550 mmCorrect answer is option 'D'. Can you explain this answer?.
A collimated beam of light of diameter 1 mm is propagating along the x- axis. The beam is to be expanded 10 mm using a combination of two convex lenses. A lens of focal length of 50 mm and another lens with focal length F are to be kept at a distance d between them. The values of F and d respectively, area)450 mm and 10 mmb)400 mm and 500 mmc)550 mm and 600 mmd)500 mm and 550 mmCorrect answer is option 'D'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A collimated beam of light of diameter 1 mm is propagating along the x- axis. The beam is to be expanded 10 mm using a combination of two convex lenses. A lens of focal length of 50 mm and another lens with focal length F are to be kept at a distance d between them. The values of F and d respectively, area)450 mm and 10 mmb)400 mm and 500 mmc)550 mm and 600 mmd)500 mm and 550 mmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A collimated beam of light of diameter 1 mm is propagating along the x- axis. The beam is to be expanded 10 mm using a combination of two convex lenses. A lens of focal length of 50 mm and another lens with focal length F are to be kept at a distance d between them. The values of F and d respectively, area)450 mm and 10 mmb)400 mm and 500 mmc)550 mm and 600 mmd)500 mm and 550 mmCorrect answer is option 'D'. Can you explain this answer?.
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