Question

g Monochromatic light with wavelength 633 nn passes through a narrow slit and a patternappears on a screen 6.0 m away. The distance on the screen between the centers of thefirst minima on either side of the screen is 32 mm. How wide (in mm) is the slit

Answer 1

Answer:

d = 1.19 x 10⁻⁴ m = 0.119 mm

Explanation:

This problem can be solved by using Young's double-slit experiment formula:

$Y = \frac{\lambda L}{d}$

where,

Y = fringe spacing = 32 mm = 0.032 m

L = slit to screen distance = 6 m

λ = wavelength of light = 633 nm = 6.33 x 10⁻⁷ m

d = slit width = ?

Therefore,

$0.032\ m = \frac{(6.33\ x\ 10^{-7}\ m)(6\ m)}{d}\\\\d = \frac{(6.33\ x\ 10^{-7}\ m)(6\ m)}{0.032\ m}$

d = 1.19 x 10⁻⁴ m = 0.119 mm