All Solutions
Page 673: Standardized Test Prep
$$
begin{align}
y = Ltan theta
end{align}
$$
where $L$ is the distance between the central bright fringe and the screen and $y$ is the vertical distance of the bright fringe from the central bright fringe.
For the wavelength, we use the condition for bringe fringes in a two-slit experiment which is given by
$$
begin{align}
dsin theta = mlambda
end{align}
$$
$$
begin{align*}
tan theta &= frac{y}{L} \ theta &= tan^{-1} left(frac{y}{L}right)
end{align*}
$$
$$
begin{align*}
theta &= tan^{-1} left(frac{2;text{mm}}{100;text{mm}}right) = boxed{1.146text{textdegree}}
end{align*}
$$
$$
begin{align*}
lambda = frac{dsin theta}{m} = (1times 10^{-6};text{m})(sin 1.146text{textdegree}) = 2.0times 10^{-8};text{m};text{or } boxed{20;text{nm}}
end{align*}
$$
$$
begin{align}
dsin theta = mlambda
end{align}
$$
We then use the linear distance from the central bright fringe relation to solve for the distance of the second bright fringe from the central bright fringe. The relationship is given by
$$
begin{align}
y = Ltan theta
end{align}
$$
where $L$ is the distance between the central bright fringe and the screen and $y$ is the vertical distance of the bright fringe from the central bright fringe.
$$
begin{align*}
sin theta &= frac{mlambda}{d} \ theta &= sin^{-1} left(frac{mlambda}{d}right)
end{align*}
$$
$$
begin{align*}
theta &= sin^{-1} left[frac{2(2times 10^{-8};text{m})}{1times 10^{-6};text{m}}right] = 2.29text{textdegree}
end{align*}
$$
$$
begin{align*}
y = (10;text{cm})(tan 2.29text{textdegree}) = 0.4;text{cm};text{or } {4;text{mm}}
end{align*}
$$
$$
begin{align}
dsin theta = mlambda
end{align}
$$
where $d$ is the slit separation distance, $theta$ is the location angle of the bright fringes, $lambda$ is the wavelength, and $m$ is the order of bright fringes.
$$
begin{align*}
sin theta = frac{mlambda}{d}
end{align*}
$$
$$
begin{align*}
2nt = mlambda quad quad m = 1,2,3,…
end{align*}
$$
where $n$ is the index of refraction of the film, $t$ is the thickness of the thin film, $lambda$ is the wavelength of the light and $m$ is the order of the dark fringe.
$$
begin{align*}
t = frac{mlambda}{2n} = frac{1(462;text{nm})}{2(1.29)} = boxed{179;text{nm}}
end{align*}
$$
$$
begin{align*}
2nt = left(m+frac{1}{2}right)lambda quad quad m = 1,2,3,…
end{align*}
$$
where $n$ is the index of refraction of the film, $t$ is the thickness of the thin film, $lambda$ is the wavelength of the light and $m$ is the order of the bright fringe.
$$
begin{align*}
lambda = frac{2nt}{m+frac{1}{2}} = frac{2(1.31)(108;text{nm})}{frac{1}{2}} = boxed{566;text{nm}}
end{align*}
$$
$$
begin{align}
y = Ltan theta
end{align}
$$
where $L$ is the distance between the central bright fringe and the screen and $y$ is the vertical distance of the bright fringe from the central bright fringe.
We use the condition for bringe fringes in a diffraction grating experiment to describe the diffraction pattern. The condition is simplified by the equation
$$
begin{align}
dsin theta = mlambda
end{align}
$$
where $d$ is the slit separation distance, $theta$ is the location angle of the bright fringes, $lambda$ is the wavelength, and $m$ is the order of bright fringes.
$$
begin{align*}
tan theta &= frac{y}{L} \ theta &= tan^{-1} left(frac{y}{L}right)
end{align*}
$$
$$
begin{align*}
theta &= tan^{-1} left(frac{54;text{mm}}{100;text{mm}}right) = 28.4text{textdegree}
end{align*}
$$
$$
begin{align*}
lambda = frac{dsin theta}{m} = (1.3times 10^{-3};text{mm})(sin 28.4text{textdegree}) = 0.476;text{mm};text{or } 476;mutext{m}
end{align*}
$$
For $lambda = 476 ;mutext{m}$,
$$
begin{align*}
sin theta = frac{mlambda}{d} = frac{1(476 ;mutext{m})}{1.3;mutext{m}} = 366
end{align*}
$$
For $lambda = 400 ;mutext{m}$,
$$
begin{align*}
sin theta = frac{mlambda}{d} = frac{1(400 ;mutext{m})}{1.3;mutext{m}} = 308
end{align*}
$$
