Question

In Fraunhofer diffraction pattern, slit width is $0.2mm$ and screen is at $2m$ away from the lens. If wavelength of light used in $5000\stackrel{\circ }{A}$ then the distance between the first minimum on either side of the central maximum is $(\theta$ is small and measure in radian):

• A. ${10}^{-1}m$
• B. ${10}^{-2}m$
• C. $2\times {10}^{-2}m$
• D. $2\times {10}^{-1}m$

Answer 1

Answer: (B)

${10}^{-2}m$

Distance between the first 2 minimum on other side $=\frac{2\lambda D}{d}$

$=\frac{2\times 5\times {10}^3\times {10}^{-10}\times 2}{2\times {10}^{-1}\times {10}^{-3}}$

$=10\times \frac{{10}^{-7}}{{10}^{-4}}$

$={10}^{-2}m$