Question

Light of wavelength $6000Angstrom$ falls on a single slit of width$0.1mm$. The second minimum will be formed for the angle of diffraction

• A. $0.06rad$
• B. $0.05rad$
• C. $0.12rad$
• D. $0.012rad$

Answer 1

The correct option is D

$0.012rad$

Step 1. Given data:

Wavelength of light $\lambda =6000Angstrom=6000\times {10}^{-10}m$

Width of slit $a=0.1mm=0.1\times {10}^{-3}m$

$n=2(secondminima)$

Step 2. Formula used:

The angle between the direction of incident light beams and any resulting diffracted beam.

The angle of the diffraction$=\frac{n\lambda }{a}$

Step 3. Calculating the angle of diffraction:

The angle of diffraction can be calculated when the angle is much smaller than $\sin \theta =\theta$

$\begin{array}{rcl}\theta & =& \frac{n\lambda }{a}\\ & =& \frac{2\times 6000\times {10}^{-10}}{0.1\times {10}^{-3}}\\ & =& 0.012rad\end{array}$

Thus, the angle of diffraction is $0·012rad$.

Hence, option D is the correct answer.