Question

In a single-slit diffraction pattern observed on a screen placed at $Dm$ distance from the slit of width $dm$, the ratio of the width of the central maximum to the width of other secondary maximum is

• A. $2:1$
• B. $1:2$
• C. $1:1$
• D. $3:1$

Answer 1

Answer: (A)

Width of central maximum
$\left(\Delta y_0\right)=2y=\frac{2\lambda D}{d}$ ...(i)
Width of 1st order secondary maxima
$=$ Distance between $D_1$ and $D_2$ (consecutive dark bands)
$=y_2-y_1$

For secondary minima (or dark band) path difference
$=d\sin \theta =m\lambda$ (where, $m=1,2,3,\dots$ )
Position of 1st dark band
Path difference $=\frac{y_{1}d}{D}=\lambda$
or $y_1=\frac{\lambda D}{d}$
Position of 2nd dark band
Path difference $=\frac{y_{2}d}{D}=2\lambda$
$\Rightarrow y_2=\frac{2\lambda D}{d}$
$\therefore$ Width of secondary maxima $\left(\Delta y_1\right)$
$\Rightarrow \Delta y_1=y_2-y_1$
or $\Delta y_1=\frac{\lambda D}{d}=y$
Thus, width of other secondary maxima is half that of central maxima.
or $\frac{\Delta y_0}{\Delta y_1}=\frac{2}{1}$