Question

The distance between the first and sixth minima in the diffraction pattern of a single slit is 0.5 mm. The screen is 0.5 m away from the slit. If the wavelength of light used is $5000A^0,$ then the slit width will be

• A. 2.5 mm
• B. 5 mm
• C. 1.25 mm
• D. 1 mm

Answer 1

The correct option is A 2.5 mm

$y_1=\left(\frac{L\lambda }{a}\right)$
$y_6=\left(\frac{6L\lambda }{a}\right)$
$y_6-y_1=\left(\frac{5L\lambda }{a}\right)$
$\left(0.5\times {10}^{-3}\right)a=5\times 5000\times {10}^{-10}\times 0.5$
$a=\left(\frac{25000\times {10}^{-10}}{{10}^{-3}}\right)$
$=25000\times {10}^{-7}$
$=2.5\times {10}^4\times {10}^{-7}$
$=2.5\times {10}^{-3}$
$a=2.5mm$