An equiconvex lens of focal length 10 cm (in air) and R.I 3/2 is put at a small opening on a tube of length 1 m fully filled with liquid of R.I. 4/3. A concave mirror of radius of curvature 20 cm is cut into two halves m1and m2and placed at the end of the tube. m1andamp; m2are placed such that their principal axes AB and CD respectively are separated by 1 mm each from the principal axes of the lens. A slit S placed in air illuminates the lens with light of frequency 7.5 andacute; 1014Hz. The light reflected from m1and m2forms interference pattern on the left end EFof the tube. O is an opaque substance to cover the gap left by m1andamp; m2.The position of the image formed by lens water combination is a andtimes; 10-2m from lens and the distance between the images formed by m1andamp; m2is b andtimes; 10-3m then findab/40..Correct answer is '8'. Can you explain this answer?

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An equiconvex lens of focal length 10 cm (in air) and R.I 3/2 is put at a small opening on a tube of length 1 m fully filled with liquid of R.I. 4/3. A concave mirror of radius of curvature 20 cm is cut into two halves m₁ and m₂ and placed at the end of the tube. m₁ & m₂ are placed such that their principal axes AB and CD respectively are separated by 1 mm each from the principal axes of the lens. A slit S placed in air illuminates the lens with light of frequency 7.5 ´ 10¹⁴ Hz. The light reflected from m₁ and m₂ forms interference pattern on the left end EFof the tube. O is an opaque substance to cover the gap left by m₁ & m₂.
The position of the image formed by lens water combination is a × 10^-2m from lens and the distance between the images formed by m₁ & m₂ is b × 10^-3 m then find ab/40.

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Correct answer is '8'. Can you explain this answer?

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The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focal length decreases. For a clear vision the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-up person.A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown-up person minimum distance of object should be around 25 cm.A person suffering from eye defects uses spectacles (Eye glass). The function of the lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina.The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with sign. For example power of lens required is +3D (converging lens of focal length 100/3 cm) then number of lens will be +3.For all the calculations required you can use the lens formula and lens makers formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens.Q.A nearsighted man can clearly see object only upto a distance of 100 cm and not beyond this. The number of the spectacles lens necessary for the remedy of this defect will be.

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The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focal length decreases. For a clear vision the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-up person.A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown-up person minimum distance of object should be around 25 cm.A person suffering from eye defects uses spectacles (Eye glass). The function of the lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina.The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with sign. For example power of lens required is +3D (converging lens of focal length 100/3 cm) then number of lens will be +3.For all the calculations required you can use the lens formula and lens makers formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens.Q.Maximum focal length of eye lens of normal person is

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The ciliary muscles of eye control the curvature of the lens in the eye and hence can alter the effective focal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focal length decreases. For a clear vision the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-up person.A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length. The ciliary muscles are most strained in this position. For an average grown-up person minimum distance of object should be around 25 cm.A person suffering from eye defects uses spectacles (Eye glass). The function of the lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina.The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with sign. For example power of lens required is +3D (converging lens of focal length100/3 cm) then number of lens will be +3.For all the calculations required you can use the lens formula and lens makers formula. Assume that the eye lens is equiconvex lens. Neglect the distance between eye lens and the spectacle lens.Q.Minimum focal length of eye lens of a normal person is

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Two thin symmetrical lenses of different nature have equal radii of curvature R = 20 cm. The lenses are placed in contact and then immersed in water. The focal length of the system is found to be 24 cm. If the refractive indices of the two lenses are 1 and 2 respectively, then find the magnitude of 9(1 - 2). Refractive index of water is 4/3. Correct answer is '5'. Can you explain this answer?

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Two thin symmetrical lenses of different nature have equal radii of curvature R = 20 cm. The lenses are placed in contact and then immersed in water. The focal length of the system is found to be 24 cm. If the refractive indices of the two lenses are 1 and 2 respectively, then find the magnitude of 9(1 - 2). Refractive index of water is 4/3. Correct answer is '5'. Can you explain this answer?

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An equiconvex lens of focal length 10 cm (in air) and R.I 3/2 is put at a small opening on a tube of length 1 m fully filled with liquid of R.I. 4/3. A concave mirror of radius of curvature 20 cm is cut into two halves m1and m2and placed at the end of the tube. m1& m2are placed such that their principal axes AB and CD respectively are separated by 1 mm each from the principal axes of the lens. A slit S placed in air illuminates the lens with light of frequency 7.5 ´ 1014Hz. The light reflected from m1and m2forms interference pattern on the left end EFof the tube. O is an opaque substance to cover the gap left by m1& m2.The position of the image formed by lens water combination is a × 10-2m from lens and the distance between the images formed by m1& m2is b × 10-3m then findab/40..Correct answer is '8'. Can you explain this answer?

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An equiconvex lens of focal length 10 cm (in air) and R.I 3/2 is put at a small opening on a tube of length 1 m fully filled with liquid of R.I. 4/3. A concave mirror of radius of curvature 20 cm is cut into two halves m1and m2and placed at the end of the tube. m1& m2are placed such that their principal axes AB and CD respectively are separated by 1 mm each from the principal axes of the lens. A slit S placed in air illuminates the lens with light of frequency 7.5 ´ 1014Hz. The light reflected from m1and m2forms interference pattern on the left end EFof the tube. O is an opaque substance to cover the gap left by m1& m2.The position of the image formed by lens water combination is a × 10-2m from lens and the distance between the images formed by m1& m2is b × 10-3m then findab/40..Correct answer is '8'. 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 An equiconvex lens of focal length 10 cm (in air) and R.I 3/2 is put at a small opening on a tube of length 1 m fully filled with liquid of R.I. 4/3. A concave mirror of radius of curvature 20 cm is cut into two halves m1and m2and placed at the end of the tube. m1& m2are placed such that their principal axes AB and CD respectively are separated by 1 mm each from the principal axes of the lens. A slit S placed in air illuminates the lens with light of frequency 7.5 ´ 1014Hz. The light reflected from m1and m2forms interference pattern on the left end EFof the tube. O is an opaque substance to cover the gap left by m1& m2.The position of the image formed by lens water combination is a × 10-2m from lens and the distance between the images formed by m1& m2is b × 10-3m then findab/40..Correct answer is '8'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An equiconvex lens of focal length 10 cm (in air) and R.I 3/2 is put at a small opening on a tube of length 1 m fully filled with liquid of R.I. 4/3. A concave mirror of radius of curvature 20 cm is cut into two halves m1and m2and placed at the end of the tube. m1& m2are placed such that their principal axes AB and CD respectively are separated by 1 mm each from the principal axes of the lens. A slit S placed in air illuminates the lens with light of frequency 7.5 ´ 1014Hz. The light reflected from m1and m2forms interference pattern on the left end EFof the tube. O is an opaque substance to cover the gap left by m1& m2.The position of the image formed by lens water combination is a × 10-2m from lens and the distance between the images formed by m1& m2is b × 10-3m then findab/40..Correct answer is '8'. Can you explain this answer?.

Solutions for An equiconvex lens of focal length 10 cm (in air) and R.I 3/2 is put at a small opening on a tube of length 1 m fully filled with liquid of R.I. 4/3. A concave mirror of radius of curvature 20 cm is cut into two halves m1and m2and placed at the end of the tube. m1& m2are placed such that their principal axes AB and CD respectively are separated by 1 mm each from the principal axes of the lens. A slit S placed in air illuminates the lens with light of frequency 7.5 ´ 1014Hz. The light reflected from m1and m2forms interference pattern on the left end EFof the tube. O is an opaque substance to cover the gap left by m1& m2.The position of the image formed by lens water combination is a × 10-2m from lens and the distance between the images formed by m1& m2is b × 10-3m then findab/40..Correct answer is '8'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of An equiconvex lens of focal length 10 cm (in air) and R.I 3/2 is put at a small opening on a tube of length 1 m fully filled with liquid of R.I. 4/3. A concave mirror of radius of curvature 20 cm is cut into two halves m1and m2and placed at the end of the tube. m1& m2are placed such that their principal axes AB and CD respectively are separated by 1 mm each from the principal axes of the lens. A slit S placed in air illuminates the lens with light of frequency 7.5 ´ 1014Hz. The light reflected from m1and m2forms interference pattern on the left end EFof the tube. O is an opaque substance to cover the gap left by m1& m2.The position of the image formed by lens water combination is a × 10-2m from lens and the distance between the images formed by m1& m2is b × 10-3m then findab/40..Correct answer is '8'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of An equiconvex lens of focal length 10 cm (in air) and R.I 3/2 is put at a small opening on a tube of length 1 m fully filled with liquid of R.I. 4/3. A concave mirror of radius of curvature 20 cm is cut into two halves m1and m2and placed at the end of the tube. m1& m2are placed such that their principal axes AB and CD respectively are separated by 1 mm each from the principal axes of the lens. A slit S placed in air illuminates the lens with light of frequency 7.5 ´ 1014Hz. The light reflected from m1and m2forms interference pattern on the left end EFof the tube. O is an opaque substance to cover the gap left by m1& m2.The position of the image formed by lens water combination is a × 10-2m from lens and the distance between the images formed by m1& m2is b × 10-3m then findab/40..Correct answer is '8'. Can you explain this answer?, a detailed solution for An equiconvex lens of focal length 10 cm (in air) and R.I 3/2 is put at a small opening on a tube of length 1 m fully filled with liquid of R.I. 4/3. A concave mirror of radius of curvature 20 cm is cut into two halves m1and m2and placed at the end of the tube. m1& m2are placed such that their principal axes AB and CD respectively are separated by 1 mm each from the principal axes of the lens. A slit S placed in air illuminates the lens with light of frequency 7.5 ´ 1014Hz. The light reflected from m1and m2forms interference pattern on the left end EFof the tube. O is an opaque substance to cover the gap left by m1& m2.The position of the image formed by lens water combination is a × 10-2m from lens and the distance between the images formed by m1& m2is b × 10-3m then findab/40..Correct answer is '8'. Can you explain this answer? has been provided alongside types of An equiconvex lens of focal length 10 cm (in air) and R.I 3/2 is put at a small opening on a tube of length 1 m fully filled with liquid of R.I. 4/3. A concave mirror of radius of curvature 20 cm is cut into two halves m1and m2and placed at the end of the tube. m1& m2are placed such that their principal axes AB and CD respectively are separated by 1 mm each from the principal axes of the lens. A slit S placed in air illuminates the lens with light of frequency 7.5 ´ 1014Hz. The light reflected from m1and m2forms interference pattern on the left end EFof the tube. O is an opaque substance to cover the gap left by m1& m2.The position of the image formed by lens water combination is a × 10-2m from lens and the distance between the images formed by m1& m2is b × 10-3m then findab/40..Correct answer is '8'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice An equiconvex lens of focal length 10 cm (in air) and R.I 3/2 is put at a small opening on a tube of length 1 m fully filled with liquid of R.I. 4/3. A concave mirror of radius of curvature 20 cm is cut into two halves m1and m2and placed at the end of the tube. m1& m2are placed such that their principal axes AB and CD respectively are separated by 1 mm each from the principal axes of the lens. A slit S placed in air illuminates the lens with light of frequency 7.5 ´ 1014Hz. The light reflected from m1and m2forms interference pattern on the left end EFof the tube. O is an opaque substance to cover the gap left by m1& m2.The position of the image formed by lens water combination is a × 10-2m from lens and the distance between the images formed by m1& m2is b × 10-3m then findab/40..Correct answer is '8'. Can you explain this answer? tests, examples and also practice JEE tests.

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