Question

In a single-slit diffraction pattern observed on a screen placed at $D m$ distance from the slit of width $d m$, 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
$\left(\Delta y_{0}\right)=2 y=\frac{2 \lambda D}{d}$ ...(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, $m=1,2,3, \ldots$ )
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 $\frac{\Delta y_{0}}{\Delta y_{1}}=\frac{2}{1}$