Single Slit Diffraction
Monochromatic light of wavelength 500 nm passes through a single slit of width 0.1 mm, so what is the angular position of the first diffraction minimum?
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About This Question
- Subject
- physics
- Chapter
- optics
- Topic
- single slit diffraction
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
Approximately 0.005 radian
In single slit diffraction, NCERT gives the condition for the dark fringes as asinθ=mλ, where a is the slit width, λ the wavelength, and m the order of the minimum. For the first minimum m=1, and for small angles sinθ≈θ, so θ≈aλ. Substituting λ=500×10−9 m and a=0.1×10−3=10−4 m gives θ≈10−4500×10−9=10−45×10−7=5×10−3 radian, or 0.005 radian. The value 0.05 radian is wrong because it misplaces the slit-width exponent by one order. The value 0.0025 radian is wrong because it halves the correct result, perhaps confusing the central maximum half-width incorrectly. The value 0.5 radian is wrong because it is far too large for such a narrow-wavelength-to-width ratio. As stated in NCERT Class 12, Chapter 10, the central maximum broadens as the slit narrows, since θ scales as λ/a. A magnitude check confirms that a slit two hundred times wider than the wavelength gives a small angle of a few milliradians, consistent with 0.005 radian.
This hard difficulty physics question is from the chapter optics, covering the topic of single slit diffraction. It appeared in the 2025 exam.
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