Answer
(a) $\frac{d}{\lambda} = 2$
(b) $\frac{d}{\lambda} = 1.15$
(c) $\frac{d}{\lambda} = 1$
Work Step by Step
We can write the general condition for a minimum:
$d~sin~\theta_m = m~\lambda$
$\frac{d}{\lambda} = \frac{m}{sin~\theta_m}$
(a) We can find the slit-to-wavelength ratio when $\theta_1 = 30^{\circ}$:
$\frac{d}{\lambda} = \frac{1}{sin~\theta_1}$
$\frac{d}{\lambda} = \frac{1}{sin~30^{\circ}}$
$\frac{d}{\lambda} = 2$
(b) We can find the slit-to-wavelength ratio when $\theta_1 = 60^{\circ}$:
$\frac{d}{\lambda} = \frac{1}{sin~\theta_1}$
$\frac{d}{\lambda} = \frac{1}{sin~60^{\circ}}$
$\frac{d}{\lambda} = 1.15$
(c) We can find the slit-to-wavelength ratio when $\theta_1 = 90^{\circ}$:
$\frac{d}{\lambda} = \frac{1}{sin~\theta_1}$
$\frac{d}{\lambda} = \frac{1}{sin~90^{\circ}}$
$\frac{d}{\lambda} = 1$
