Question

In a Young's double-slit experiment the wavelength of light used is 466 nm (in vacuum), and the separation between the slits is 1.3 × 10^-6 m. Determine the angle that locates: (a) the dark fringe for which m = 0 (b) the bright fringe for which m = 1 (c) the dark fringe for which m = 1 and (d) the bright fringe for which m = 2.

Answer 1

Answer:

Part a)

$\theta_1 = 10.3 degree$

Part b)

$\theta_2 = 21 degree$

Part c)

$\theta_3 = 32.5 degree$

Part d)

$\theta_4 = 45.8 degree$

Explanation:

For the position of dark fringe the path difference of light is odd multiple of half of the wavelength

so here we will have

$dsin\theta = \frac{2m + 1}{2}\lambda$

part a)

for m = 0 we have

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

$(1.3 \times 10^{-6})sin\theta = \frac{466 \times 10^{-9}}{2}$

$\theta = 10.3 degree$

Part b)

Similarly for the maximum intensity we will have path difference must be integral multiple of wavelength

so we have

$d sin\theta = N\lambda$

for m = 1 we have

$(1.3 \times 10^{-6})sin\theta = 466 \times 10^{-9}$

$\theta = 21 degree$

Part c)

for m = 1 we have

$d sin\theta = \frac{3\lambda}{2}$

$(1.3 \times 10^{-6})sin\theta = \frac{3(466 \times 10^{-9})}{2}$

$\theta = 32.5 degree$

Part d)

for m = 2 again we have

$d sin\theta = N\lambda$

$(1.3 \times 10^{-6})sin\theta = 2\times 466 \times 10^{-9}$

$\theta = 45.8 degree$