Question

In a single slit diffraction pattern-the distance between the first minimum on the left and the first minimum on the right is 5 mm. The Screen on which the diffraction pattern is displayed at a distance of 80 cm from the slit The length is $6000\stackrel{\text{o}}{A}$ .The slit width in (mm) is about:

• A. $0.576$
• B. $0.348$
• C. $0.192$
• D. $0.096$
• E. $0.048$

Answer 1

Answer: (C)

Slit width, $W=\frac{D\lambda }{d}$ Given, $D=80cm=80\times {10}^{-2}m,$ $\lambda =6000\mathring{A}=6000\times {10}^{-10}m,$ $d=\frac{5}{2}mm=\frac{5\times {10}^{-3}}{2}m$ $\therefore$ $W=\frac{80\times {10}^{-2}\times 6000\times {10}^{-10}\times 2}{5\times {10}^{-3}}$ $W=0.192\times {10}^{-3}m=0.192mm$