Question

Consider a single slit that produces a first-order minimum at 16.5° when illuminated with monochromatic light. show answer No Attempt 50% Part (a) At what angle, in degrees, is the second-order minimum?

Answer 1

Answer:

$\theta_2 = 34.61^0$

Explanation:

Path difference for the destructive interference of a single slit:

$D sin \theta = n \lambda$

For the first - order minimum, n = 1, and $\theta = \theta_1$

$D sin \theta_1 = \lambda$.........(1)

For the second - order minimum, n = 2,  and $\theta = \theta_2$

$D sin \theta_2 = 2 \lambda$.........(2)

Dividing equation (2) by equation (1):

$\frac{D sin \theta_2}{Dsin \theta_1} = \frac{2 \lambda}{\lambda} \\\frac{ sin \theta_2}{sin \theta_1} = 2 \\\theta_1 = 16.5^0\\\frac{ sin \theta_2}{sin 16.5} = 2\\sin \theta_2 = 2 sin 16.5\\sin \theta_2 = 0.568\\\theta_2 = sin^{-1} 0.568\\\theta_2 = 34.61^0$