Question

Light of wavelength 500 nm falls on a single slit of width 0.1 mm. The angular position of the first minimum is}

• A. \(\sin^{-1}(0.05)\)
• B. \(\sin^{-1}(0.2)\)
• C. \(\sin^{-1}(0.5)\)
• D. \(\sin^{-1}(0.005)\)
• E. \(\sin^{-1}(0.0025)\)

Hint:

For first minimum in single slit diffraction: \[ a\sin\theta=\lambda \] Always convert wavelength and slit width into SI units first.

Answer 1

The Correct Option is D

For single slit diffraction, first minimum occurs at: \[ a\sin\theta = \lambda \] Given: \[ \lambda = 500\text{ nm}=5\times 10^{-7}\text{ m} \] \[ a=0.1\text{ mm}=10^{-4}\text{ m} \] So, \[ \sin\theta=\frac{\lambda}{a} =\frac{5\times 10^{-7}}{10^{-4}} =5\times 10^{-3} \] \[ \sin\theta=0.005 \] Thus, \[ \theta=\sin^{-1}(0.005) \]
Hence, the correct answer is: \[ \boxed{(D)\ \sin^{-1}(0.005)} \]