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Violet light is incident on a double slit. The distance between the two slits is 0.040mm, and aninterference pattern is created ona screen that is 2.0m away from the slits. If the 5th bright fringe on the screen is 10.0cm away from the central fringe, what is the wavelength of the light?

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Answer:

The wavelength of the light is
0.04 * 10^(-5) m

Step-by-step explanation:

Given :

Distance between two slit
d = 0.040 * 10^(-3) m

Distance of screen
D = 2 m

Distance between central fringe to 5th bright fringe,
x = 10 * 10^(-2) m

From the formula of interference,


d \sin \theta = n \lambda

Where
\sin \theta
\theta we put
\theta = (x)/(D)


(dx)/(D) = n\lambda

Where
n = 5 ( given in question )

Now we have to find wavelength of the light,


\lambda = (dx)/(nD)


\lambda = (0.040 * 10^(-3) * 10 * 10^(-2) )/(5 * 2)


\lambda = 0.04 * 10^(-5) m

Therefore, the wavelength of the light is
0.04 * 10^(-5) m

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