Question

Light of wavelength 465 nm passes through a single slit of width 2.32*10^-5 m. At what angle does the first interference minimum (m=1) occur?

Answer 1

Answer: 1.15 is right i tried it on accllus its for (m=1) it is right not wrong

Step-by-step explanation:

Answer 2

1.15°

Step-by-step explanation:

The width of slit, angle of interference and wavelength are related with the formula

`wsin\theta=m\lambda`

and making

`\theta` the subject of formula we get

`\theta=sin^(-1)(\frac {m\lambda}{w})`

Where

`\theta` is the angle where the first interference occurs, w is width,

`\lambda` is the wavelength

Substituting 1 for m,
`232*10^(-5) m` for w and
`465*10^(-9) m` for
`\lambda`

`\theta=sin^(-1)(\frac {1*465* 10^(-9)}{232* 10^(-5)})=1.14846213920053^(\circ)`

Therefore, rounded off, the angle is approximately 1.15°