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Angular width of a central max. is $30^o$ when the slit is illuminated by light of wavelength 6000 Å. Then width of the slit will be approx. |
12 × 10–6 m 12 × 10–7 m 12 × 10–8 m 12 × 10–9 m |
12 × 10–7 m |
$\text{Angular width } 2\theta = 30^o$ $sin 30^o = \frac{\lambda}{a}$ $a = \frac{\lambda}{sin30^o} = 2\lambda$ $ a = 12000A^o = 12\times 10^{-7}m$ |
