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You are given a converging lens with equal radii of curvature and a diverging lens with the same radii of curvature as those of the converging lens. The lenses are made of material with n = 1.50, and the radii of curvature are all 35 cm. They are placed at opposite ends of a tube 15 cm long, and the nearer lens which is converging is 10 cm from an object. What is the location of the image that results from the two refractions? Write the final distance (in cm) from diverging lens.
Correct answer is between '15.85,15.87'. Can you explain this answer?
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You are given a converging lens with equal radii of curvature and a di...
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You are given a converging lens with equal radii of curvature and a di...
Given:
- Converging lens and diverging lens with equal radii of curvature (35 cm)
- Material of the lenses has refractive index n = 1.50
- Lenses are placed at opposite ends of a tube 15 cm long
- The converging lens is 10 cm from an object
To find:
- The location of the image resulting from the two refractions
- The final distance (in cm) from the diverging lens
Solution:
1. Calculating the focal length of the lenses:
- The focal length of a lens is given by the formula: f = R / (n - 1), where R is the radius of curvature and n is the refractive index
- For both lenses, R = 35 cm and n = 1.50
- Therefore, the focal length of both lenses is f = 35 cm / (1.50 - 1) = 35 cm / 0.50 = 70 cm
2. Calculating the position of the image formed by the converging lens:
- Using the lens formula: 1/f = 1/v - 1/u, where v is the image distance and u is the object distance
- For the converging lens, f = 70 cm and u = -10 cm (since the object is 10 cm in front of the lens)
- Substitute the values into the lens formula: 1/70 = 1/v - 1/-10
- Simplify the equation: 1/v = 1/70 + 1/10 = (1 + 7)/70 = 8/70
- Take the reciprocal of both sides: v = 70/8 = 8.75 cm
- Therefore, the image formed by the converging lens is located at a distance of 8.75 cm from the lens.
3. Calculating the position of the final image formed by the diverging lens:
- Since the lenses are placed at opposite ends of a tube, the image formed by the converging lens acts as the object for the diverging lens.
- The distance between the two lenses is 15 cm, and the image formed by the converging lens is located 8.75 cm from the converging lens.
- Therefore, the object distance for the diverging lens is 15 cm - 8.75 cm = 6.25 cm (measured from the diverging lens)
- Using the lens formula again, but this time for the diverging lens: 1/f = 1/v - 1/u
- For the diverging lens, f = -70 cm (since it is a diverging lens) and u = -6.25 cm
- Substitute the values into the lens formula: 1/-70 = 1/v - 1/-6.25
- Simplify the equation: 1/v = -1/70 + 1/6.25 = (-6.25 + 70)/(70 * 6.25) = 63.75/437.5 = 0.1457
- Take the reciprocal of both sides: v = 1/0.1457 = 6.868 cm
- Therefore, the final image formed by the diver
- Converging lens and diverging lens with equal radii of curvature (35 cm)
- Material of the lenses has refractive index n = 1.50
- Lenses are placed at opposite ends of a tube 15 cm long
- The converging lens is 10 cm from an object
To find:
- The location of the image resulting from the two refractions
- The final distance (in cm) from the diverging lens
Solution:
1. Calculating the focal length of the lenses:
- The focal length of a lens is given by the formula: f = R / (n - 1), where R is the radius of curvature and n is the refractive index
- For both lenses, R = 35 cm and n = 1.50
- Therefore, the focal length of both lenses is f = 35 cm / (1.50 - 1) = 35 cm / 0.50 = 70 cm
2. Calculating the position of the image formed by the converging lens:
- Using the lens formula: 1/f = 1/v - 1/u, where v is the image distance and u is the object distance
- For the converging lens, f = 70 cm and u = -10 cm (since the object is 10 cm in front of the lens)
- Substitute the values into the lens formula: 1/70 = 1/v - 1/-10
- Simplify the equation: 1/v = 1/70 + 1/10 = (1 + 7)/70 = 8/70
- Take the reciprocal of both sides: v = 70/8 = 8.75 cm
- Therefore, the image formed by the converging lens is located at a distance of 8.75 cm from the lens.
3. Calculating the position of the final image formed by the diverging lens:
- Since the lenses are placed at opposite ends of a tube, the image formed by the converging lens acts as the object for the diverging lens.
- The distance between the two lenses is 15 cm, and the image formed by the converging lens is located 8.75 cm from the converging lens.
- Therefore, the object distance for the diverging lens is 15 cm - 8.75 cm = 6.25 cm (measured from the diverging lens)
- Using the lens formula again, but this time for the diverging lens: 1/f = 1/v - 1/u
- For the diverging lens, f = -70 cm (since it is a diverging lens) and u = -6.25 cm
- Substitute the values into the lens formula: 1/-70 = 1/v - 1/-6.25
- Simplify the equation: 1/v = -1/70 + 1/6.25 = (-6.25 + 70)/(70 * 6.25) = 63.75/437.5 = 0.1457
- Take the reciprocal of both sides: v = 1/0.1457 = 6.868 cm
- Therefore, the final image formed by the diver
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You are given a converging lens with equal radii of curvature and a diverging lens with the same radii of curvature as those of the converging lens. The lenses are made of material with n = 1.50, and the radii of curvature are all 35 cm. They are placed at opposite ends of a tube 15 cm long, and the nearer lens which is converging is 10 cm from an object. What is the location of the image that results from the two refractions? Write the final distance (in cm) from diverging lens.Correct answer is between '15.85,15.87'. Can you explain this answer?
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You are given a converging lens with equal radii of curvature and a diverging lens with the same radii of curvature as those of the converging lens. The lenses are made of material with n = 1.50, and the radii of curvature are all 35 cm. They are placed at opposite ends of a tube 15 cm long, and the nearer lens which is converging is 10 cm from an object. What is the location of the image that results from the two refractions? Write the final distance (in cm) from diverging lens.Correct answer is between '15.85,15.87'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about You are given a converging lens with equal radii of curvature and a diverging lens with the same radii of curvature as those of the converging lens. The lenses are made of material with n = 1.50, and the radii of curvature are all 35 cm. They are placed at opposite ends of a tube 15 cm long, and the nearer lens which is converging is 10 cm from an object. What is the location of the image that results from the two refractions? Write the final distance (in cm) from diverging lens.Correct answer is between '15.85,15.87'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for You are given a converging lens with equal radii of curvature and a diverging lens with the same radii of curvature as those of the converging lens. The lenses are made of material with n = 1.50, and the radii of curvature are all 35 cm. They are placed at opposite ends of a tube 15 cm long, and the nearer lens which is converging is 10 cm from an object. What is the location of the image that results from the two refractions? Write the final distance (in cm) from diverging lens.Correct answer is between '15.85,15.87'. Can you explain this answer?.
You are given a converging lens with equal radii of curvature and a diverging lens with the same radii of curvature as those of the converging lens. The lenses are made of material with n = 1.50, and the radii of curvature are all 35 cm. They are placed at opposite ends of a tube 15 cm long, and the nearer lens which is converging is 10 cm from an object. What is the location of the image that results from the two refractions? Write the final distance (in cm) from diverging lens.Correct answer is between '15.85,15.87'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about You are given a converging lens with equal radii of curvature and a diverging lens with the same radii of curvature as those of the converging lens. The lenses are made of material with n = 1.50, and the radii of curvature are all 35 cm. They are placed at opposite ends of a tube 15 cm long, and the nearer lens which is converging is 10 cm from an object. What is the location of the image that results from the two refractions? Write the final distance (in cm) from diverging lens.Correct answer is between '15.85,15.87'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for You are given a converging lens with equal radii of curvature and a diverging lens with the same radii of curvature as those of the converging lens. The lenses are made of material with n = 1.50, and the radii of curvature are all 35 cm. They are placed at opposite ends of a tube 15 cm long, and the nearer lens which is converging is 10 cm from an object. What is the location of the image that results from the two refractions? Write the final distance (in cm) from diverging lens.Correct answer is between '15.85,15.87'. Can you explain this answer?.
Solutions for You are given a converging lens with equal radii of curvature and a diverging lens with the same radii of curvature as those of the converging lens. The lenses are made of material with n = 1.50, and the radii of curvature are all 35 cm. They are placed at opposite ends of a tube 15 cm long, and the nearer lens which is converging is 10 cm from an object. What is the location of the image that results from the two refractions? Write the final distance (in cm) from diverging lens.Correct answer is between '15.85,15.87'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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You are given a converging lens with equal radii of curvature and a diverging lens with the same radii of curvature as those of the converging lens. The lenses are made of material with n = 1.50, and the radii of curvature are all 35 cm. They are placed at opposite ends of a tube 15 cm long, and the nearer lens which is converging is 10 cm from an object. What is the location of the image that results from the two refractions? Write the final distance (in cm) from diverging lens.Correct answer is between '15.85,15.87'. Can you explain this answer?, a detailed solution for You are given a converging lens with equal radii of curvature and a diverging lens with the same radii of curvature as those of the converging lens. The lenses are made of material with n = 1.50, and the radii of curvature are all 35 cm. They are placed at opposite ends of a tube 15 cm long, and the nearer lens which is converging is 10 cm from an object. What is the location of the image that results from the two refractions? Write the final distance (in cm) from diverging lens.Correct answer is between '15.85,15.87'. Can you explain this answer? has been provided alongside types of You are given a converging lens with equal radii of curvature and a diverging lens with the same radii of curvature as those of the converging lens. The lenses are made of material with n = 1.50, and the radii of curvature are all 35 cm. They are placed at opposite ends of a tube 15 cm long, and the nearer lens which is converging is 10 cm from an object. What is the location of the image that results from the two refractions? Write the final distance (in cm) from diverging lens.Correct answer is between '15.85,15.87'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice You are given a converging lens with equal radii of curvature and a diverging lens with the same radii of curvature as those of the converging lens. The lenses are made of material with n = 1.50, and the radii of curvature are all 35 cm. They are placed at opposite ends of a tube 15 cm long, and the nearer lens which is converging is 10 cm from an object. What is the location of the image that results from the two refractions? Write the final distance (in cm) from diverging lens.Correct answer is between '15.85,15.87'. Can you explain this answer? tests, examples and also practice JEE tests.
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