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White light is incident on a glass plate of thickness 5000 and refractive index 1.5. The wavelength in the visible region (4000 Å - 7000 Å) that is strongly reflected by the plate is (a) 4290 A (b) 8000 Å (c) 4000 A (d) 5000?
Most Upvoted Answer
White light is incident on a glass plate of thickness 5000 and refract...
For maximum intensity,
Δx=2μt=(2n−1)2λ
⇒λ=2n−14μt
=2n−14×1.5×500
=2n−13000nm
substituting 1,2,3.... we get
λ=3000nm,1000nm,600nm
So the required answer is 600 nm.
Δx=2μt=(2n−1)2λ
⇒λ=2n−14μt
=2n−14×1.5×500
=2n−13000nm
substituting 1,2,3.... we get
λ=3000nm,1000nm,600nm
So the required answer is 600 nm.
Community Answer
White light is incident on a glass plate of thickness 5000 and refract...
Calculation of Wavelength Strongly Reflected by Glass Plate
White light incident on a glass plate of thickness 5000 Å and refractive index 1.5 will undergo reflection and refraction.
Determination of Reflected Wavelength
- When light is incident on a medium with a higher refractive index, some of it gets reflected at the interface. The reflected light will be in phase if the optical path length difference between the two rays is an integral multiple of the wavelength.
- The phase change upon reflection from a denser medium is π radians. So, for constructive interference to occur, the optical path difference should be an odd multiple of half the wavelength.
- The condition for constructive interference is given by 2nt = (2m + 1)λ, where t is the thickness of the glass plate, n is the refractive index, m is an integer, and λ is the wavelength.
- Substituting the given values, we get 2 * 1.5 * 5000 = (2m + 1)λ.
- Solving the equation, we find λ = 4290 Å.
Therefore, the wavelength in the visible region that is strongly reflected by the glass plate is 4290 Å, option (a).
White light incident on a glass plate of thickness 5000 Å and refractive index 1.5 will undergo reflection and refraction.
Determination of Reflected Wavelength
- When light is incident on a medium with a higher refractive index, some of it gets reflected at the interface. The reflected light will be in phase if the optical path length difference between the two rays is an integral multiple of the wavelength.
- The phase change upon reflection from a denser medium is π radians. So, for constructive interference to occur, the optical path difference should be an odd multiple of half the wavelength.
- The condition for constructive interference is given by 2nt = (2m + 1)λ, where t is the thickness of the glass plate, n is the refractive index, m is an integer, and λ is the wavelength.
- Substituting the given values, we get 2 * 1.5 * 5000 = (2m + 1)λ.
- Solving the equation, we find λ = 4290 Å.
Therefore, the wavelength in the visible region that is strongly reflected by the glass plate is 4290 Å, option (a).
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White light is incident on a glass plate of thickness 5000 and refractive index 1.5. The wavelength in the visible region (4000 Å - 7000 Å) that is strongly reflected by the plate is (a) 4290 A (b) 8000 Å (c) 4000 A (d) 5000?
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White light is incident on a glass plate of thickness 5000 and refractive index 1.5. The wavelength in the visible region (4000 Å - 7000 Å) that is strongly reflected by the plate is (a) 4290 A (b) 8000 Å (c) 4000 A (d) 5000? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about White light is incident on a glass plate of thickness 5000 and refractive index 1.5. The wavelength in the visible region (4000 Å - 7000 Å) that is strongly reflected by the plate is (a) 4290 A (b) 8000 Å (c) 4000 A (d) 5000? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for White light is incident on a glass plate of thickness 5000 and refractive index 1.5. The wavelength in the visible region (4000 Å - 7000 Å) that is strongly reflected by the plate is (a) 4290 A (b) 8000 Å (c) 4000 A (d) 5000?.
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