Question

Light of wavelength 687 nm is incident on a single slit 0.75 mm wide. At what distance from the slit should a screen be placed if the second dark fringe in the diffraction pattern is to be 1.7 mm from the center of the diffraction pattern?

Answer 1

Answer:

L = 0.93 meter

Explanation:

As we know that the diffraction pattern will show the minimum on the screen at the position where it follows the equation

$a sin\theta = N\lambda$

here we know that for second fringe position we have

$sin\theta = \frac{2\lambda}{a}$

$sin\theta = \frac{2(687 nm)}{0.75 mm}$

$sin\theta = 1.832 \times 10^{-3}$

$\theta = 0.105 degree$

now we know that the position of minimum is given at 1.7 mm from the center

so we have

$tan\theta = \frac{y}{L}$

$tan(0.105) = \frac{1.7\times 10^{-3}}{L}$

$L = 0.93 meter$