Question:
In Young's double slit experiment, slit separation is 0.6 mm and the separation between slit and screen is 1.2 m. The angular width is (the wavelength of light used is 4800 \( \dot{A} \))
In Young's double slit experiment, slit separation is 0.6 mm and the separation between slit and screen is 1.2 m. The angular width is (the wavelength of light used is 4800 \( \dot{A} \))
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In Young's double slit experiment, slit separation is 0.6 mm and the separation between slit and screen is 1.2 m. The angular width is (the wavelength of light used is 4800 $\dotA$)
Updated On: Apr 15, 2026
- 30 rad
- $8\times10^{-4}$ rad
- 12 rad
- 70.5 rad
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The Correct Option is B
Solution and Explanation
Step 1: Formula
Angular width $\theta$ is given by $\theta = \frac{\lambda}{d}$.
Step 2: Values
$\lambda = 4800 \times 10^{-10} \text{ m}$ and $d = 0.6 \text{ mm} = 0.6 \times 10^{-3} \text{ m}$.
Step 3: Calculation
$\theta = \frac{4800 \times 10^{-10}}{0.6 \times 10^{-3}} = 8000 \times 10^{-7} = 8 \times 10^{-4} \text{ rad}$.
Final Answer: (B)
Angular width $\theta$ is given by $\theta = \frac{\lambda}{d}$.
Step 2: Values
$\lambda = 4800 \times 10^{-10} \text{ m}$ and $d = 0.6 \text{ mm} = 0.6 \times 10^{-3} \text{ m}$.
Step 3: Calculation
$\theta = \frac{4800 \times 10^{-10}}{0.6 \times 10^{-3}} = 8000 \times 10^{-7} = 8 \times 10^{-4} \text{ rad}$.
Final Answer: (B)
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