Question

Light of wavelength 656 nm and 410 nm emitted from a hot gas of hydrogen atoms strikes a grating with 5300 lines per centimeter. a) Determine the angular deflection of both wavelengths in the 1st and 2nd order.

Answer 1

Answer:

$20.32^{\circ}$ and $44.08^{\circ}$

$12.56^{\circ}$ and $25.77^{\circ}$

Explanation:

$\lambda$ = Wavelength

$\theta$ = Angle

m = Order

Distance between grating is given by

$d=\dfrac{1}{5300}\\\Rightarrow d=0.0001886\ \text{cm}$

$\lambda=656\ \text{nm}$

We have the relation

$d\sin\theta=m\lambda\\\Rightarrow \theta=\sin^{-1}\dfrac{m\lambda}{d}$

m = 1

$\theta=\sin^{-1}\dfrac{1\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=20.35^{\circ}$

m = 2

$\theta=\sin^{-1}\dfrac{2\times 656\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=44.08^{\circ}$

The first and second order angular deflection is $20.32^{\circ}$ and $44.08^{\circ}$

$\lambda=410\ \text{nm}$

m = 1

$\theta=\sin^{-1}\dfrac{1\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=12.56^{\circ}$

m = 2

$\theta=\sin^{-1}\dfrac{2\times 410\times 10^{-9}}{0.0001886\times 10^{-2}}\\\Rightarrow \theta=25.77^{\circ}$

The first and second order angular deflection is $12.56^{\circ}$ and $25.77^{\circ}$.