Question

A parallel beam of light of wavelength $4000\stackrel{°}{A}$ passes through a slit of width $5\times {10}^{-3}m$. The angular spread of the central maxima in the diffraction pattern is

• A. $1.6\times {10}^{-3}rad$
• B. $1.6\times {10}^{-4}rad$
• C. $3.2\times {10}^{-3}rad$
• D. $3.2\times {10}^{-4}rad$

Answer 1

The correct option is B

$1.6\times {10}^{-4}rad$

Step 1: Given data

The wavelength of the given light, $\lambda =4000\stackrel{°}{A}$

Slit width, $a=5\times {10}^{-3}m$

Step 2: Formula Used

The half angular width of central maxima,

$sin\theta =\frac{\lambda }{a}$

Step 3: Calculation

The half angular width of central maxima,

$$\begin{gathered} sin\theta =\frac{\lambda }{a} \\ \Rightarrow \theta =\frac{\lambda }{a}(\because \sin \theta \approx \theta ) \end{gathered}$$

Angular spread of the central maxima in the diffraction pattern

$$\begin{gathered} =2\theta \\ =\frac{2\lambda }{a} \\ =\frac{2\times 4000\times {10}^{-10}}{5\times {10}^{-3}} \\ =1.6\times {10}^{-4}rad \end{gathered}$$

Therefore, the angular spread of the central maxima in the diffraction pattern is $1.6\times {10}^{-4}rad$.

Option B is correct.