Question

light of wavelength 610 nm is incident on a narrow slit. The angle between the first diffraction minimum on one side of the central meximum and the first minimum on the other side is 1.23°. What is the width of the slit?

Answer 1

The width of the slit is
`d = 5.68 *10^(-5) \ m`

Step-by-step explanation:

From the question we are told that

The wavelength is
`\lambda = 610 \ nm = 610 *10^(-9) \ m`

The angle is
`\theta = 1.23 ^o`

Generally the angle between the first minimum on one side and that the central maximum is evaluated as

`\theta _1 = (\theta)/(2)`

=>
`\theta _1 = (1.23)/(2)`

=>
`\theta _1 = 0.615 ^o`

Generally the condition for minimum diffraction is mathematically represented as

`d sin \theta_1 = n\lambda`

For first minimum n = 1

=>
`d = (n \lambda )/( sin (\theta_1))`

=>
`d = (1 * 610 *10^(-9))/( sin (0.615))`

=>
`d = 5.68 *10^(-5) \ m`