Question

A diffraction grafting experiment is set up using orange light of wavelength $600nm.$
The grafting has a slit separation of $2.0\mu m.$
What is the angular separation $\left( {{\theta _2} - {\theta _1}} \right)$ between the first and second order maxima of the orange light?
(A) ${17.5^0}$
(B) ${19.4^0}$
(C) ${36.9^0}$
(D) ${54.3^0}$

Answer 1

Hint : The diffraction grating is an optical component with a periodic structure which splits and diffracts light into the several beams travelling in the different directions. Use formula, $n\lambda = d\sin \theta$ where, $n - 1,2,3,....$ and $\lambda$ is the wavelength, $\theta =$ the angle of emergence and d is the slit separation.

Complete Step By Step Answer:
Given that the wavelength, $\lambda = 600nm$
Place the value of nano-metre -
$\lambda = 600 \times {10^{ - 9}}m$
$d = 2\mu m \\\ d = 2 \times {10^{ - 6}}m \\\$
Now using the equation of the diffraction grating,
$n\lambda = d\sin \theta$
$\sin \theta = \dfrac{{n\lambda }}{d}$
When the angle is very small,
We can write,
${\theta _n} \approx \dfrac{{n\lambda }}{d}$
Place the value of $n = 1,2$
${\theta _2} = \dfrac{{2\lambda }}{d}{\text{ }}.........{\text{(1)}}$
Similarly,
${\theta _1} = \dfrac{\lambda }{d}{\text{ }}.........{\text{(2)}}$
Subtract the equation $(2)$ from $(1)$
${\theta _2} - {\theta _1} = \dfrac{{2\lambda }}{d}{\text{ - }}\dfrac{\lambda }{d}{\text{ }} \\\ {\theta _2} - {\theta _1} = \dfrac{\lambda }{d}{\text{ }} \\\$
Place the given values in the above equations –
${\theta _2} - {\theta _1} = \dfrac{{600 \times {{10}^{ - 9}}}}{{2 \times {{10}^{ - 6}}}}$
According to the property when the negative powers are moved from the denominator to the numerator or vice-versa the powers are changed to positive.
${\theta _2} - {\theta _1} = \dfrac{{600 \times {{10}^6}}}{{2 \times {{10}^9}}} \\\ {\theta _2} - {\theta _1} = \dfrac{{300 \times {{10}^6}}}{{{{10}^9}}} \\\ {\theta _2} - {\theta _1} = \dfrac{3}{{{{10}^{}}}} \\\ {\theta _2} - {\theta _1} = 0.3\,{\text{radian}} \\\$
Convert radians into degree –
$1{\text{ Degree = }}\dfrac{{{\text{180}}^\circ }}{\pi } \times {\text{radians}}$
Therefore,
$0.3{\text{ radians = }}\dfrac{{{\text{180}}}}{{3.1415}} \times 0.3 \\\ 0.3{\text{ radians }} = 17.5^\circ \\\$
Therefore, the required solution - A diffraction grafting experiment is set up using orange light of wavelength $600nm.$ and the grafting has a slit separation of $2.0\mu m.$ then, the angular separation $\left( {{\theta _2} - {\theta _1}} \right)$ between the first and second order maxima of the orange light is $17.5^\circ$
Hence, from the given multiple choices, option A is the correct answer.

Note :
Always check the given units and convert them in the same format. As the same we converted nano-meters and micrometers in metres. Follow the different system of formats to solve these types of examples. There are three systems of formats to describe units.
A) MKS System (Meter Kilogram System )
B.) CGS System (Centimetre Gram System)
C.) SI Unit System International