Question

A plane wave of wavelength $6250\AA$is incident normally on a slit of width $2\times {10}^{-2}cm$. The width of the principal maximum on a screen distant $50cm$ will be

• A. $312.5\times {10}^{-3}cm$
• B. $31.25\times {10}^{-3}cm$
• C. $312cm$
• D. $312.5\times {10}^{-3}mm$

Answer 1

The correct option is A

$312.5\times {10}^{-3}cm$

Step 1: Given data

Wavelength of a plane wave, $\lambda =6250\AA$

$$\begin{gathered} =6250\times {10}^{-10}\times {10}^{2}cm \\ =6250\times {10}^{-8}cm \end{gathered}$$

Width of a slit, $a=2\times {10}^{-2}cm$

Distance between the screen and slit, $D=50cm$

Step 2: Find the angular width of the central maxima, $\mathit{\beta }$

The angular width of central maxima, $\beta =2\theta =2(\lambda /a)$

Step 3: Find the linear width of the central maxima, $\mathit{L}$

Linear width of the central maxima, $L=\beta \times D$

$$\begin{gathered} =(2\lambda /a)\times D \\ =\left[\right(2\times 6250\times {10}^{-8})/(2\times {10}^{-2}\left)\right]\times 50 \\ =3125\times {10}^{-4}cm \\ =312.5\times {10}^{-3}cm \end{gathered}$$

Hence, option (A) is correct.