For Question 1: A light of wavelength 610 nm produces a first maximum at an angle of 30 degrees. What is the slit separation? For Question 2: 3rd minimum is observed at 30 degrees for a slit separatio

The wavelength of light is $\lambda = 610 \text{ nm}$.

$$\lambda = 610 \times 10^{-9} \text{ m}$$

The order of the maximum is $m = 1$ (first maximum).
The angle is $\theta = 30^\circ$.

The condition for a maximum is given by:

$$d \sin \theta = m \lambda$$

where $d$ is the slit separation.

$$d \sin(30^\circ) = 1 \times (610 \times 10^{-9} \text{ m})$$

We know that $\sin(30^\circ) = 0.5$.

$$d \times 0.5 = 610 \times 10^{-9} \text{ m}$$ $$d = \frac{610 \times 10^{-9} \text{ m}}{0.5}$$ $$d = 1220 \times 10^{-9} \text{ m}$$ $$d = 1.22 \times 10^{-6} \text{ m}$$

This can also be expressed as $1.22 \text{ µm}$.

The separation between the slits is $\boxed{\text{1.22 \text{ \textmu}m}}$.

---

The order of the minimum is $m = 3$ (3rd minimum).
The angle is $\theta = 30^\circ$.
The slit separation is $d = 3 \text{ cm}$.

$$d = 3 \times 10^{-2} \text{ m}$$

The condition for a minimum is given by:

$$d \sin \theta = \left(m - \frac{1}{2}\right) \lambda$$

where $\lambda$ is the wavelength of light.

$$(3 \times 10^{-2} \text{ m}) \sin(30^\circ) = \left(3 - \frac{1}{2}\right) \lambda$$

We know that $\sin(30^\circ) = 0.5$.

$$(3 \times 10^{-2} \text{ m}) \times 0.5 = \left(\frac{5}{2}\right) \lambda$$ $$1.5 \times 10^{-2} \text{ m} = 2.5 \lambda$$ $$\lambda = \frac{1.5 \times 10^{-2} \text{ m}}{2.5}$$ $$\lambda = 0.6 \times 10^{-2} \text{ m}$$ $$\lambda = 6 \times 10^{-3} \text{ m}$$

This wavelength is very large for visible light. Let's recheck the calculation.

$$\lambda = 0.006 \text{ m}$$

This is $6 \text{ mm}$. This is an unusually large wavelength for light. However, based on the given values and formula, this is the calculated result.

The wavelength of light is $\boxed{\text{6 \text{ mm}}}$.