Question

In Young’s double slit experiment with sodium vapour lamp of wavelength 589 nm and slit 0.589 mm apart, the half angular width of the central maxima is

• A. \(\sin^{-1}(0.01)\)
• B. \(\sin^{-1}(0.0001)\)
• C. \(\sin^{-1}(0.001)\)
• D. \(\sin^{-1}(0.1)\)

Hint:

Always convert wavelength and slit separation to the same units before calculating their ratio in interference problems.

Answer 1

The Correct Option is C

In Young’s double slit experiment, the angular position of the central maxima is given by: \[ \theta = \sin^{-1} \left( \frac{\lambda}{d} \right) \] Given: - Wavelength, \( \lambda = 589 \, \text{nm} = 589 \times 10^{-9} \, \text{m} \) - Slit separation, \( d = 0.589 \, \text{mm} = 0.589 \times 10^{-3} \, \text{m} \) Now calculate: \[ \frac{\lambda}{d} = \frac{589 \times 10^{-9}}{0.589 \times 10^{-3}} = 0.001 \] Hence: \[ \theta = \sin^{-1}(0.001) \]