Question

way from the slits. If the 7th bright fringe on the screen is a linear distance of 10.0 cm away from the central fringe, what is the wavelength of the light? In a two-slit experiment, the slit separation is 3.00 × 10-5 m. The interference pattern is created on a screen that is 2.00 m away from the slits. If the 7th bright fringe on the screen is a linear distance of 10.0 cm away from the central fringe, what is the wavelength of the light? 214 nm 204 nm 224 nm 100 nm 234 nm

Answer 1

Answer:

The value is  $\lambda = 214.3 \ nm$

Explanation:

From the question we are told that

   The  slit separation is  $d = 3.00 * 10^{-5} m$

    The  distance of the screen is  $D = 2.00\ m$

    The  order of fringe is  n  =  7

    The path difference is  $y = 10.0 \ cm = 0.1 \ m$

    Generally the path difference is mathematically represented as

      $y = \frac{n * \lambda * D}{ d}$

=>   $0.1 = \frac{7 * \lambda * 2.00 }{ 3.00 * 10^{-5}}$

=>  $\lambda = \frac{0.1 *3.00 * 10^{-5} }{7 * 2.00 }$

=>   $\lambda = \frac{0.1 *3.00 * 10^{-5} }{7 * 2.00 }$

=>   $\lambda = 2.143 *10^{-7} \ m$    

=>    $\lambda = 214.3 \ nm$