Suppose that the central peak of a single-slit diffraction pattem is so wide that the first minima can be assumed to occur at angular positions of \(\pm 90^{\circ} .\) For this case, what is the ratio of the slit width to the wavelength of the light?

Understanding diffraction minima is crucial when studying light behavior as it encounters obstacles like slits. Diffraction refers to the bending and spreading out of waves around the edges of an obstacle or through a slit. The single-slit diffraction pattern produced shows a central bright peak, flanked on either side by dark bands called minima. These minima occur at specific angles where the light waves interfere destructively. To be more specific, they cancel each other out, resulting in areas with no light intensity.

As per the principle governing single-slit diffraction, the condition for minima is given by the equation \(a \sin \theta_n = n\lambda\), where \(a\) represents the width of the slit, \(\lambda\) the wavelength of the light, \(\theta_n\) the angle of the nth minimum, and \(n\) an integer known as the order of the minimum. The complete cancelation of waves at the first minimum is particularly interesting; here, \(\theta_n = 90^\circ\), which leads to a fundamental understanding of how the slit width is directly proportional to the wavelength of the light used.

In any experiment involving diffraction, the ratio of the slit width to the wavelength \(\frac{a}{\lambda}\) offers significant insight into the behavior of the pattern formed. For instance, in our problem, where the first minimum occurs at \(\theta_n = 90^\circ\), this ratio determines the size of the central peak.

If the slit width is much greater than the wavelength, \(a \gg \lambda\), the resulting pattern will exhibit a narrow central peak with closely spaced minima. Conversely, if \(a \approx \lambda\), as in the hypothetical scenario given where the ratio is 1:1, the central peak becomes very wide - so wide, in fact, that the first minima occur at \(\pm 90^\circ\). This special case tells us that the slit allows the wave to spread out at its maximum possible angle due to the slit's width being comparable to the wavelength of the light.

The angle of the diffraction minimum, \(\theta_n\), is an angular measure that indicates where the light intensity falls to zero in the diffraction pattern due to destructive interference. This angle can be calculated using the equation mentioned earlier: \(a \sin \theta_n = n\lambda\).

For the specific case in the exercise, the angle of the first diffraction minimum is given as \(\pm 90^\circ\). When \(n = 1\), corresponding to the first minimum, and \(\sin 90^\circ = 1\), the equation simplifies considerably, showing that the slit width equals the wavelength. This implies that when the angle of the diffraction minimum is at its theoretical maximum (\(90^\circ\)), we encounter a unique situation where the phenomenon of diffraction completely dictates the pattern, leading to an intuitive understanding that the spreading angle of the wave is fundamentally related to the dimensions of the slit and the wavelength of the wave itself.