Question

12. Most light emitted by hot sodium atoms has a wavelength 3893 nm. When passed through a grating, the first-order angula deflection of this light is 16.1°. (a) Calculate the separation adjacent lines in the grating and the number of lines per centim ter. (b) Calculate the angular deflection of the third-order lin (c) Calculate the distance on a screen located 2.0 m from the gra ing of the first-order line from the central maximum.

Answer 1

Given:

$wavelength \lambda =589.3 nm=589.3\times10^{-9}m$

$Angular deflection\Theta _1=16.1 degree$

Part a:

Let d be the separation of adjacent line in the grating.

Using the formula;

Here m=1 for first order angular deflection

$dsin\Theta =m\lambda \\d=\frac{m\lambda }{sin\Theta } \\d=\frac{589.3\times10^{-9}}{sin16.1\degree} \\d=2125\times10^{-9}m$

Number of lines per meter = 1/ d = 4.7 * 10⁵

Number of line per centimeter = 4.7 * 10³

Part b:

Angular deflection of the third order line :

$\Theta _3=arcsin(\frac{3\lambda }{d}) \\\Theta _3=arcsin\left \lfloor \frac{3\times589.3\times10^{-9}}{2125\times10^{-9}} \right \rfloor \\\Theta _3=56.3\degree$

Part c:

Let the distance on the screen of the of the first order line is y.

$sin\Theta _1=\frac{y}{D} \\y=Dsin\Theta _1 \\y=2sin16.1\degree \\y=0.555m$