Question

The first diffraction minimum due to a single slit Fraunhoffer diffraction is at the angle of diffraction ${30}^{\circ }$ for a light of wave length 5000 $A^0$ .The width of the slit is

• A. $1.0\times {10}^{-4}cm$
• B. $2.164\times {10}^{-4}cm$
• C. $1.082\times {10}^{-3}cm$
• D. $0.546cm$

Answer 1

The correct option is A $1.0\times {10}^{-4}cm$

$\begin{array}{c}Accordingtoquestion...............\\ ForDiffractionminima=\\ d\sin \theta =n\lambda \\ 1stordern=1\\ \Rightarrow dsin30=1\times 5000\times {10}^{-10}\\ \Rightarrow d\times \frac{1}{2}=5000\times {10}^{-10}\\ \Rightarrow d=10000\times {10}^{-10}m\\ \therefore d=1\times {10}^{-4}m\\ sothatthecorrectoptionisA.\end{array}$