**Problem 6.91: What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at $\theta\:=\: 45.0^0$ ?**

**Solution:**

In diffraction $n^th$ order minima is obtained at an angle given by $sin\:\theta\:=\:\frac{n\lambda}{a}$ where $\lambda$ is the wavelength of light and $a$ is the width of slit.

From this ratio of the slit width to the wavelength, $\frac{\lambda}{a}\:=\:\frac{sin\:\theta}{n}$.

For the first diffraction minimum $n$ is equal to $1$.

Therefore, $\frac{\lambda}{a}\:=\:sin\:\theta\:=\:sin\:\theta\:=\:1$