Question

How wide is the central diffraction peak on a screen 2.20 mm behind a 0.0328-mmmm-wide slit illuminated by 588-nmnm light?

Answer 1

Answer:

$y = 0.0394 \ m$

Explanation:

From the question we are told that

        The  distance of the screen is  $D = 2.20 \ m$

       The distance of separation of the slit is  $d = 0.0328 \ mm = 0.0328*10^{-3} \ m$

        The  wavelength of light is  $\lambda = 588 \ nm = 588 *10^{-9} \ m$

Generally the condition for constructive interference is

            $dsin\theta = n * \lambda$

=>        $\theta = sin^{-1} [ \frac{ n * \lambda }{d } ]$

here n = 1 because we are considering the central diffraction peak

=>        $\theta = sin^{-1} [ \frac{ 1 * 588*10^{-9} }{0.0328*10^{-3} } ]$

=>       $\theta = 1.0274 ^o$

Generally the width of central diffraction peak on a screen is mathematically evaluated as

           $y = D tan (\theta )$

substituting values

        $y = 2.20 * tan (1.0274)$

        $y = 0.0394 \ m$