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Consider light that has its third minimum at an angle of 23.3° when it falls on a single slit of width 4.05 μm. Find the wavelength of the light in nanometers.

User Brezotom
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1 Answer

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Answer:

The wavelength of light is 533 nm.

Step-by-step explanation:

It is given that,

Width of a single slit,
d=4.05* 10^(-6)\ m

Light has its third minimum at an angle of 23.3° when it falls on a single slit. For destructive interference, the equation for minima is given by:


d\ sin\theta=n\lambda

Here, n = 3


\lambda=(d\ sin\theta)/(n)


\lambda=(4.05* 10^(-6)* sin(23.3))/(3)


\lambda=5.33* 10^(-7)\ m


\lambda=533\ nm

So, the wavelength of the light is 533 nm. Hence, this is the required solution.

User Lauraann
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