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Collimated beam of light of diameter 1 mm is propagating along the x-axis. The beam is to be expanded to a collimated beam of diameter 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 value of F and d respectively are?
Most Upvoted Answer
Collimated beam of light of diameter 1 mm is propagating along the x-a...
Given:
- Diameter of initial beam = 1 mm
- Diameter of final beam = 10 mm
- Focal length of first lens = 50 mm
To find:
- Focal length of second lens (F)
- Distance between the two lenses (d)
Explanation:
To expand the collimated beam of light from 1 mm to 10 mm, we need to use a combination of two convex lenses. Let's solve this step by step.
Step 1: Calculating the focal length of the second lens (F)
The formula for lens combination is:
1/F = 1/f1 + 1/f2
We are given the focal length of the first lens (f1) as 50 mm. We can substitute these values in the formula and solve for F.
1/F = 1/50 + 1/f2
Simplifying further, we get:
1/f2 = 1/F - 1/50
Now, we need to find the focal length of the second lens (f2) when the diameter of the beam is expanded to 10 mm. To do this, we can use the lens formula:
1/f2 = (1/v) - (1/u)
Where v is the image distance and u is the object distance. Since the beam is collimated, the object distance (u) will be infinity. Therefore, we can ignore the 1/u term in the equation.
1/f2 = 1/v
Now, using the lens formula, we can write:
1/f2 = 1/v = 1/F - 1/50
Simplifying further, we get:
1/f2 = 1/F - 1/50
Therefore, the focal length of the second lens (F) is 50 mm.
Step 2: Calculating the distance between the two lenses (d)
To find the distance between the two lenses, we can use the lens formula for the first lens:
1/f1 = (1/v1) - (1/u1)
Since the beam is collimated, the object distance (u1) will be infinity. Therefore, we can ignore the 1/u1 term in the equation.
1/f1 = 1/v1
Now, substituting the values, we get:
1/50 = 1/v1
Simplifying further, we get:
v1 = 50 mm
The image formed by the first lens (Lens 1) will act as the object for the second lens (Lens 2). Therefore, the distance between the two lenses (d) will be equal to the image distance (v1) formed by the first lens.
Therefore, the distance between the two lenses is 50 mm.
Final Answer:
The focal length of the second lens (F) is 50 mm and the distance between the two lenses (d) is also 50 mm.
- Diameter of initial beam = 1 mm
- Diameter of final beam = 10 mm
- Focal length of first lens = 50 mm
To find:
- Focal length of second lens (F)
- Distance between the two lenses (d)
Explanation:
To expand the collimated beam of light from 1 mm to 10 mm, we need to use a combination of two convex lenses. Let's solve this step by step.
Step 1: Calculating the focal length of the second lens (F)
The formula for lens combination is:
1/F = 1/f1 + 1/f2
We are given the focal length of the first lens (f1) as 50 mm. We can substitute these values in the formula and solve for F.
1/F = 1/50 + 1/f2
Simplifying further, we get:
1/f2 = 1/F - 1/50
Now, we need to find the focal length of the second lens (f2) when the diameter of the beam is expanded to 10 mm. To do this, we can use the lens formula:
1/f2 = (1/v) - (1/u)
Where v is the image distance and u is the object distance. Since the beam is collimated, the object distance (u) will be infinity. Therefore, we can ignore the 1/u term in the equation.
1/f2 = 1/v
Now, using the lens formula, we can write:
1/f2 = 1/v = 1/F - 1/50
Simplifying further, we get:
1/f2 = 1/F - 1/50
Therefore, the focal length of the second lens (F) is 50 mm.
Step 2: Calculating the distance between the two lenses (d)
To find the distance between the two lenses, we can use the lens formula for the first lens:
1/f1 = (1/v1) - (1/u1)
Since the beam is collimated, the object distance (u1) will be infinity. Therefore, we can ignore the 1/u1 term in the equation.
1/f1 = 1/v1
Now, substituting the values, we get:
1/50 = 1/v1
Simplifying further, we get:
v1 = 50 mm
The image formed by the first lens (Lens 1) will act as the object for the second lens (Lens 2). Therefore, the distance between the two lenses (d) will be equal to the image distance (v1) formed by the first lens.
Therefore, the distance between the two lenses is 50 mm.
Final Answer:
The focal length of the second lens (F) is 50 mm and the distance between the two lenses (d) is also 50 mm.
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Collimated beam of light of diameter 1 mm is propagating along the x-axis. The beam is to be expanded to a collimated beam of diameter 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 value of F and d respectively are?
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Collimated beam of light of diameter 1 mm is propagating along the x-axis. The beam is to be expanded to a collimated beam of diameter 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 value of F and d respectively are? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Collimated beam of light of diameter 1 mm is propagating along the x-axis. The beam is to be expanded to a collimated beam of diameter 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 value of F and d respectively are? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Collimated beam of light of diameter 1 mm is propagating along the x-axis. The beam is to be expanded to a collimated beam of diameter 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 value of F and d respectively are?.
Collimated beam of light of diameter 1 mm is propagating along the x-axis. The beam is to be expanded to a collimated beam of diameter 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 value of F and d respectively are? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about Collimated beam of light of diameter 1 mm is propagating along the x-axis. The beam is to be expanded to a collimated beam of diameter 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 value of F and d respectively are? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Collimated beam of light of diameter 1 mm is propagating along the x-axis. The beam is to be expanded to a collimated beam of diameter 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 value of F and d respectively are?.
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Collimated beam of light of diameter 1 mm is propagating along the x-axis. The beam is to be expanded to a collimated beam of diameter 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 value of F and d respectively are?, a detailed solution for Collimated beam of light of diameter 1 mm is propagating along the x-axis. The beam is to be expanded to a collimated beam of diameter 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 value of F and d respectively are? has been provided alongside types of Collimated beam of light of diameter 1 mm is propagating along the x-axis. The beam is to be expanded to a collimated beam of diameter 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 value of F and d respectively are? theory, EduRev gives you an
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