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Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed on a screen that is 2.30m from the grating.In the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm. What is the difference between these wavelengths?

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Answer:


\Delta \lambda=14.3\ nm

Step-by-step explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm,
\Delta Y=2.98\ mm = 0.00298\ m

Let d is the slit width of the grating,


d=(1)/(N)


d=(1)/(900\ cm)


d=1.11* 10^(-5)\ m

For the first wavelength, the position of maxima is given by :


y_1=(L\lambda_1)/(d)

For the other wavelength, the position of maxima is given by :


y_2=(L\lambda_2)/(d)

So,


\Delta \lambda=(\Delta y d)/(l)


\Delta \lambda=(0.00298* 1.11* 10^(-5))/(2.3)


\Delta \lambda=1.43* 10^(-8)\ m

or


\Delta \lambda=14.3\ nm

So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.

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