A Level Physics Practice Exam

In a single slit diffraction pattern, how does the size of dark fringes change with a decrease in wavelength?

1. Dark fringes become larger

2. Dark fringes become smaller

3. Dark fringes disappear

4. Dark fringes remain the same size

The correct answer is: Dark fringes become smaller

In a single slit diffraction pattern, the dark fringes are formed due to the destructive interference of light waves that pass through the slit. The location and size of these dark fringes are influenced by the wavelength of the light being used. As the wavelength decreases, the angle at which dark fringes occur also changes according to the formula for destructive interference in a single slit, which can be expressed as: \[ a \sin(\theta) = m \lambda \] where \(a\) is the width of the slit, \(m\) is the order of the dark fringe (m=1,2,3,...), and \(\lambda\) is the wavelength of the light. When the wavelength \(\lambda\) decreases, for a fixed slit width \(a\), the values of \(\sin(\theta)\) corresponding to each dark fringe (for each order \(m\)) will also decrease, meaning that the angles at which the dark fringes appear become smaller. This geometric effect implies that the spacing between the dark fringes reduces, leading to the conclusion that as the wavelength decreases, the dark fringes get smaller. In essence, a decrease in wavelength results in more closely spaced dark fringes, causing them