Q.46. In single slit diffraction pattern, the wavelength of light used is 628 nm and slit width is 0

Q: What is the correct answer for this exam Physics Q46?

The correct answer is 36. Sol. The angular width of the central maximum in a single-slit diffraction pattern is given by $\theta = \frac{2\lambda}{d}$, where $\lambda$ is the wavelength and $d$ is the slit width. Given $\lambda = 628 \text{ nm} = 628 \times 10^{-9} \text{ m}$ and $d = 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m}$. So $\theta = \frac{2 \times 628 \times 10^{-9}}{0.2 \times 10^{-3}} = \frac{1256 \times 10^{

Q: Q.46. 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 degrees. The value of is _______.

The correct answer is 36. Sol. The angular width of the central maximum in a single-slit diffraction pattern is given by $\theta = \frac{2\lambda}{d}$, where $\lambda$ is the wavelength and $d$ is the slit width. Given $\lambda = 628 \text{ nm} = 628 \times 10^{-9} \text{ m}$ and $d = 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m}$. So $\theta = \frac{2 \times 628 \times 10^{-9}}{0.2 \times 10^{-3}} = \frac{1256 \times 10^{

Question

Q.46. 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  degrees. The value of  is _______.

FirstInTest · Accepted Answer

Sol. The angular width of the central maximum in a single-slit diffraction pattern is given by $	heta = \frac{2\lambda}{d}$, where $\lambda$ is the wavelength and $d$ is the slit width. Given $\lambda = 628 	ext{ nm} = 628 	imes 10^{-9} 	ext{ m}$ and $d = 0.2 	ext{ mm} = 0.2 	imes 10^{-3} 	ext{ m}$. So $	heta = \frac{2 	imes 628 	imes 10^{-9}}{0.2 	imes 10^{-3}} = \frac{1256 	imes 10^{-9}}{0.2 	imes 10^{-3}} = 6280 	imes 10^{-6} = 6.28 	imes 10^{-3} 	ext{ radians}$. To convert radians to degrees, multiply by $\frac{180}{\pi}$. $	heta_{degrees} = 6.28 	imes 10^{-3} 	imes \frac{180}{\pi} \approx 6.28 	imes 10^{-3} 	imes \frac{180}{3.14} = 6.28 	imes 10^{-3} 	imes 57.32 \approx 0.36 	ext{ degrees}$. The problem states the angular width is $\alpha 	imes 10^{-2}$ degrees. So $\alpha 	imes 10^{-2} = 0.36 \Rightarrow \alpha = 36$.

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