Question

36.4 diffraction by a double slit
if we increase the wavelength of the light used to form a double - slit diffraction pattern,

the width of the central diffraction peak increases and the number of bright fringes within the peak stays the same.

the width of the central diffraction peak decreases and the number of bright fringes within the peak decreases.

the width of the central diffraction peak decreases and the number of bright fringes within the peak increases.

the width of the central diffraction peak increases and the number of bright fringes within the peak increases.

the width of the central diffraction peak increases and the number of bright fringes within the peak decreases.

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Explanation:

1. Recall the formula for the width of the central diffraction peak in double - slit diffraction: The angular width of the central diffraction peak is given by $\Delta\theta = \frac{2\lambda}{a}$, where $\lambda$ is the wavelength of light and $a$ is the width of each slit. When the wavelength $\lambda$ increases (while $a$ remains constant), the angular width $\Delta\theta$ of the central diffraction peak increases. This means the width of the central diffraction peak (both angular and linear, if we consider a screen at a fixed distance) increases.
2. Recall the formula for the position of the bright fringes in double - slit interference: The position of the $m^{th}$ bright fringe is given by $y_m=\frac{m\lambda L}{d}$, where $L$ is the distance between the slits and the screen, $d$ is the distance between the two slits, and $m = 0,\pm1,\pm2,\cdots$. The number of bright fringes within the central diffraction peak is related to the ratio of the interference fringe spacing and the diffraction peak width. The condition for a bright fringe to be within the central diffraction peak is that $|m|\leq\frac{a}{d}$. Since $\lambda$ does not appear in this condition (the condition is based on the ratio of the slit width $a$ and the slit separation $d$), the number of bright fringes within the central diffraction peak stays the same when we only change the wavelength $\lambda$.

Answer:

the width of the central diffraction peak increases and the number of bright fringes within the peak stays the same.