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A parallel beam of sodium light of wavelength 5890 A is incident on a thin glass plate of refractive index 1.5 such that the angle of refraction in the plate is 60degree. The smallest thickness of the plate which will make it dark by reflection is (A) 5980 A (B) 7856 A (C) 1964 A (D) 3928 A?
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A parallel beam of sodium light of wavelength 5890 A is incident on a ...
The given problem involves the phenomenon of thin film interference. Let's break down the solution into several steps:

1. Determine the incident angle: The angle of refraction in the glass plate is given as 60 degrees. Since the glass plate is thin, we can assume the angle of incidence is also 60 degrees.

2. Use Snell's Law to find the refractive index of the glass: Snell's Law states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive indices of the two media. In this case, we have:
sin(60°) / sin(angle of refraction in glass) = 1 / refractive index of glass
sin(60°) / sin(60°) = 1 / refractive index of glass
1 = 1 / refractive index of glass
refractive index of glass = 1.5

3. Calculate the path length difference: When light reflects from the top and bottom surfaces of the glass plate, a path length difference is created. For destructive interference (dark fringe), this path length difference must be equal to an odd multiple of half the wavelength. Since the incident light is a parallel beam, the path length difference is simply twice the thickness of the glass plate.

4. Set up the interference condition: The path length difference is given by 2 * thickness. To make the glass plate dark by reflection, the path length difference should be equal to (2n + 1) * λ/2, where n is an integer and λ is the wavelength of the sodium light in the glass.

5. Solve for the thickness: Putting the values into the equation:
2 * thickness = (2n + 1) * λ/2
2 * thickness = (2n + 1) * 5890 A/2
thickness = (2n + 1) * 5890 A/4
For the smallest thickness, we take n = 0:
thickness = (2 * 0 + 1) * 5890 A/4
thickness = 5890 A/4
thickness = 1472.5 A

6. Convert to the appropriate units: The thickness is given in Angstroms. To match the answer choices, we need to convert it to A.

7. Compare the answer choices: The calculated thickness is 1472.5 A, which is closest to option (C) 1964 A.

Therefore, the correct answer is option (C) 1964 A.
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A parallel beam of sodium light of wavelength 5890 A is incident on a thin glass plate of refractive index 1.5 such that the angle of refraction in the plate is 60degree. The smallest thickness of the plate which will make it dark by reflection is (A) 5980 A (B) 7856 A (C) 1964 A (D) 3928 A?
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A parallel beam of sodium light of wavelength 5890 A is incident on a thin glass plate of refractive index 1.5 such that the angle of refraction in the plate is 60degree. The smallest thickness of the plate which will make it dark by reflection is (A) 5980 A (B) 7856 A (C) 1964 A (D) 3928 A? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A parallel beam of sodium light of wavelength 5890 A is incident on a thin glass plate of refractive index 1.5 such that the angle of refraction in the plate is 60degree. The smallest thickness of the plate which will make it dark by reflection is (A) 5980 A (B) 7856 A (C) 1964 A (D) 3928 A? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A parallel beam of sodium light of wavelength 5890 A is incident on a thin glass plate of refractive index 1.5 such that the angle of refraction in the plate is 60degree. The smallest thickness of the plate which will make it dark by reflection is (A) 5980 A (B) 7856 A (C) 1964 A (D) 3928 A?.
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