All Solutions
Page 664: Lesson Check
$$
begin{align*}
m &= 1 \
theta &= 1.25 text{textdegree} \
lambda &= 587.5 text{ nm}
end{align*}
$$
The split spacing for the diffraction grating to the first order principal maximum of the yellow light from a helium discharge tube can be obtain by applying the expression for constructive interference by a diffraction grating :
$$
begin{align*}
d sin theta &= m lambda \
d &= dfrac{m lambda}{sin theta} \
&= dfrac{1 (587.5 cdot 10^{-9} text{ m})}{sin 1.25 text{textdegree}}
end{align*}
$$
$$
{boxed{d = 26.93 ~ mu text{m}}}
$$
{d = 26.93 ~ mu text{m}}
$$
$$
begin{align*}
m &= 2 \
d &= 1.92 cdot 10^{-6} text{ m} \
lambda &= 692 text{ nm}
end{align*}
$$
The angle to the second order principal maximum of a light on a diffraction grating can be obtain by applying the expression for constructive interference by a diffraction grating :
$$
begin{align*}
d sin theta &= m lambda \
sin theta &= dfrac{m lambda}{d} \
theta &= {sin}^{-1} left( dfrac{m lambda}{d} right) \
&= {sin}^{-1} left( dfrac{2 (692 cdot 10^{-9} text{ m})}{1.92 cdot 10^{-6} text{ m}} right)
end{align*}
$$
$$
{boxed{theta = 46.12 text{textdegree}}}
$$
{theta = 46.12 text{textdegree}}
$$
$$
begin{align*}
m &= 1 \
d &= 2.2 ~ mu text{m} \
theta &= 21 text{textdegree}
end{align*}
$$
The wavelength of the light that shines on a diffraction grating through an angle to the first order principal maximum can be obtain by applying the expression for constructive interference by a diffraction grating :
$$
begin{align*}
d sin theta &= m lambda \
lambda &= dfrac{d sin theta}{m} \
&= dfrac{2.2 cdot 10^{-6} text{ m} (sin 21 text{textdegree})}{1}
end{align*}
$$
$$
{boxed{lambda = 788.41 text{ nm}}}
$$
{lambda = 788.41 text{ nm}}
$$