JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2026
- Medium
- Numerical
In single slit diffraction pattern, the wavelength of light used is \(628\) nm and slit width is \(0.2\) mm, the angular width of central maximum is \(\alpha \times 10^{-2}\) degrees. The value of \(\alpha\) is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 36
Step-by-step Solution
Detailed explanation
The angular width of the central maximum in a single slit diffraction pattern is given by \(\Delta \theta = \dfrac{2\lambda}{a}\)
Substituting the given values:
\(\lambda = 628 \times 10^{-9}\) m
\(a = 0.2 \times 10^{-3}\) m
\(\Delta \theta = \dfrac{2 \times 628 \times 10^{-9}}{0.2 \times 10^{-3}}\) rad
\(\Delta \theta = 6280 \times 10^{-6}\) rad \(= 6.28 \times 10^{-3}\) rad
To convert radians to degrees, multiply by \(\dfrac{180}{\pi}\):
\(\Delta \theta = 6.28 \times 10^{-3} \times \dfrac{180}{3.14}\) degrees
\(\Delta \theta = 2 \times 10^{-3} \times 180\) degrees
\(\Delta \theta = 360 \times 10^{-3}\) degrees \(= 36 \times 10^{-2}\) degrees
Comparing this with \(\alpha \times 10^{-2}\) degrees, we get \(\alpha = 36\).
Answer: \(36\)
Substituting the given values:
\(\lambda = 628 \times 10^{-9}\) m
\(a = 0.2 \times 10^{-3}\) m
\(\Delta \theta = \dfrac{2 \times 628 \times 10^{-9}}{0.2 \times 10^{-3}}\) rad
\(\Delta \theta = 6280 \times 10^{-6}\) rad \(= 6.28 \times 10^{-3}\) rad
To convert radians to degrees, multiply by \(\dfrac{180}{\pi}\):
\(\Delta \theta = 6.28 \times 10^{-3} \times \dfrac{180}{3.14}\) degrees
\(\Delta \theta = 2 \times 10^{-3} \times 180\) degrees
\(\Delta \theta = 360 \times 10^{-3}\) degrees \(= 36 \times 10^{-2}\) degrees
Comparing this with \(\alpha \times 10^{-2}\) degrees, we get \(\alpha = 36\).
Answer: \(36\)
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