Question

(a) at what angle is the first minimum for 550-nm light falling on a single slit of width 1.00μm1.00μm ? (b) will there be a second minimum?

Answer 1

The required angled is$\theta=33.4^{\circ}$.

What is angle is the first minimum?

The first minimum is going to be theta one is the inverse sine of one times 550 times ten to the minus nine meters divided by one times ten to the minus six meters giving an angle of 33.4 degrees

$\begin{aligned}& \lambda=550 \mathrm{~nm} \\& D=1.00 \mathrm{~nm}\end{aligned}$

$D \sin \theta=m \lambda, m=1,-1,2,-2, \ldots$

a) $\theta=\sin ^{-1}\left(\frac{m \lambda}{D}\right)$

$\theta_1=\sin ^{-1}\left(\frac{(1)\left(550 \times 10^{-9} \mathrm{~m}\right)}{1.00 \times 10^{-6} \mathrm{~m}}\right)=33.4^{\circ}$

b)The value of sine function can never be greater than 1, so second minimum does not exists for given wavelength of light and width of slit.

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