Question

QUESTIONS You use a diffraction grating with 3000 lines / cm to separate two spectral lines in hydrogen, with wavelengths of 485 nm and 656 nm. If the screen to observe the diffraction pattern is 15 cm from the gratine what is the distance on the screen between the first order maxima for the two spectral lines? Enter your answer in cm. Worning: The small angle approximation is not valid in this question

Answer 1

The grating has 3000 lines/ cm. That is 3000 lines in 1 cm

So, the distance between the lines in the grating is

1 1 d = 10-2m 3000 = 3.33 x 10-6, 3000/ cm 3000/10-2m

The condition for constructive interference for a diffraction grating (also known as diffraction grating equation) is

dsin(0) = ml
  for m= 0, 1, −1, 2, −2
where
d is the distance between the lines in the grating
$\lambda$ is the wavelength of the light

$\theta$ is the angle at which maixima occur for a given order and a given wavelength
and m is the order of the maximum

The diffraction grating considered in the figure below produces the two wavelengths
TV
and $\lambda_{2}$ on the screen which is located at a distance x=15 cm from the grating. Point A denotes the point at which first order maxima for the

wavelength
TV
occurs and point B denotes the point at which first order maxima for the wavelength  $\lambda_{2}$ occurs. $\theta_{1}$ and $\theta_{2}$ are the angles made by the emergent rays w.r.t the horizontal (x). y1 and y2 denote the perpendicular distance of each of the first order maxima from the center and $\Delta y$ is the distance between the first order maxima for the two spectral lines on the screen.

B K Y2 TOL x=15cm Grating Screen

In order to find the distance between the first order maxima for the two spectral lines, we find the angle $\theta$ for each wavelength using the diffraction equation,

a) for $\lambda_{1}=486 nm=486\times 10^{-9}m$

$dsin(\theta_{1})=m\lambda_{1}$

for first order maxima, m=1

$\therefore dsin(\theta_{1})=\lambda_{1}$

Rearranging,

$sin(\theta_{1})=\frac{\lambda_{1}}{d}=\frac{486\times 10^{-9}}{3.33\times 10^{-6}}=0.1459$

$\theta_{1}=sin^{-1}(0.1459)=8.39\degree$

b) for $\lambda_{2}=656 nm=656\times 10^{-9}m$

$dsin(\theta_{2})=m\lambda_{2}$

for first order maxima, m=1

$\therefore dsin(\theta_{2})=\lambda_{2}$

Rearranging,

$sin(\theta_{2})=\frac{\lambda_{2}}{d}=\frac{656\times 10^{-9}}{3.33\times 10^{-6}}=0.1969$

A2 = sin (0.1969) = 11.36°

Now we have to calculate the distance y1 and y2 ,

From figure,

$tan(\theta_{1})=\frac{y_{1}}{x}$

Rearranging,

$y_{1}=xtan(\theta_{1})$

$\therefore y_{1}=(15\times 10^{-2})tan(8.39)=0.022m$

Similary ,

$tan(\theta_{2})=\frac{y_{2}}{x}$

Rearranging,

$y_{2}=xtan(\theta_{2})$

$\therefore y_{2}=(15\times 10^{-2})tan(11.36)=0.030m$

Now the distance between the first order maxima for the two spectral lines on the screen $=\Delta y=y_{2}-y_{1}$

$\therefore \Delta y=0.030-0.022=0.008m=8 \times 10^{-3}m=8mm$

is the required distance