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

`\theta_2 = 34.61^0`

Step-by-step 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):

`(D sin \theta_2)/(Dsin \theta_1) = (2 \lambda)/(\lambda) \\( sin \theta_2)/(sin \theta_1) = 2 \\\theta_1 = 16.5^0\\( 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`