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A thin oil film of refractive index 1.2 floats on the surface of water ( m=4/3 ) . When a light of wavelength l = 9.6 × 10–7 m falls normally on the film from air, then it appears dark when seen normally. The minimum change in its thickness for which it will appear bright in normally reflected light by the same light is:
- a)10-7 m
- b)2 x 10-7 m
- c)3 x 10-7 m
- d)5 x 10-7 m
Correct answer is option 'B'. Can you explain this answer?
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A thin oil film of refractive index 1.2 floats on the surface of water...
For normal incidence path difference between ray 1 and ray 2 is 2μ1 t
For minimum thickness increment 
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Most Upvoted Answer
A thin oil film of refractive index 1.2 floats on the surface of water...
Μm falls normally on the film, calculate the minimum thickness of the film for constructive interference to occur.
Given:
Refractive index of oil (n1) = 1.2
Refractive index of water (n2) = 4/3
Wavelength of light (λ) = 9.6 μm
For constructive interference to occur, the path difference between the reflected rays from the top and bottom surfaces of the film should be an integer multiple of the wavelength.
Let the thickness of the film be t. The path difference is given by:
2nt = mλ
Where:
n is the refractive index of the medium (in this case, it is n1 for the oil film)
m is an integer (1, 2, 3,...)
Since the light reflects off both the top and bottom surfaces of the film, the total path difference is twice the thickness of the film multiplied by the refractive index of the oil.
2n1t = mλ
Substitute the values of n1, λ, and solve for t:
2(1.2)t = λ
2(1.2)t = 9.6 μm
2.4t = 9.6
t = 9.6 / 2.4
t = 4 μm
Therefore, the minimum thickness of the oil film for constructive interference to occur is 4 μm.
Given:
Refractive index of oil (n1) = 1.2
Refractive index of water (n2) = 4/3
Wavelength of light (λ) = 9.6 μm
For constructive interference to occur, the path difference between the reflected rays from the top and bottom surfaces of the film should be an integer multiple of the wavelength.
Let the thickness of the film be t. The path difference is given by:
2nt = mλ
Where:
n is the refractive index of the medium (in this case, it is n1 for the oil film)
m is an integer (1, 2, 3,...)
Since the light reflects off both the top and bottom surfaces of the film, the total path difference is twice the thickness of the film multiplied by the refractive index of the oil.
2n1t = mλ
Substitute the values of n1, λ, and solve for t:
2(1.2)t = λ
2(1.2)t = 9.6 μm
2.4t = 9.6
t = 9.6 / 2.4
t = 4 μm
Therefore, the minimum thickness of the oil film for constructive interference to occur is 4 μm.
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A thin oil film of refractive index 1.2 floats on the surface of water ( m=4/3 ).When a light of wavelengthl = 9.6 × 10–7 m falls normally on the film from air, then it appears dark when seen normally. The minimum change in its thickness for which it will appear bright in normally reflected light by the same light is: a)10-7 mb)2 x10-7 mc)3 x10-7 md)5 x10-7 mCorrect answer is option 'B'. Can you explain this answer?
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A thin oil film of refractive index 1.2 floats on the surface of water ( m=4/3 ).When a light of wavelengthl = 9.6 × 10–7 m falls normally on the film from air, then it appears dark when seen normally. The minimum change in its thickness for which it will appear bright in normally reflected light by the same light is: a)10-7 mb)2 x10-7 mc)3 x10-7 md)5 x10-7 mCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A thin oil film of refractive index 1.2 floats on the surface of water ( m=4/3 ).When a light of wavelengthl = 9.6 × 10–7 m falls normally on the film from air, then it appears dark when seen normally. The minimum change in its thickness for which it will appear bright in normally reflected light by the same light is: a)10-7 mb)2 x10-7 mc)3 x10-7 md)5 x10-7 mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A thin oil film of refractive index 1.2 floats on the surface of water ( m=4/3 ).When a light of wavelengthl = 9.6 × 10–7 m falls normally on the film from air, then it appears dark when seen normally. The minimum change in its thickness for which it will appear bright in normally reflected light by the same light is: a)10-7 mb)2 x10-7 mc)3 x10-7 md)5 x10-7 mCorrect answer is option 'B'. Can you explain this answer?.
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