Find the angular fringe width in Young’s double slits experiment with a blue-green light of wavelength of $6000{A}^{°}$. The separation between the slits is $3 \times {10}^{-3}m$.

Hint: Young’s double slit experiment was the first experiment to observe the interference of light. So, to solve this problem, use the formula for angular fringe width for interference in Young’s double slit experiment. The formula for angular fringe width used is in terms of separation between the slits and wavelength of the light. Substitute the values in the above-mentioned formula and find the angular fringe width.

Formula used:
$\beta= \dfrac {\lambda}{d}$

Complete answer:
Given: Wavelength of blue-green light, $\lambda =6000{A}^{°}=6000 \times {10}^{-10} m$
            Separation between the slits, $d=3 \times {10}^{-3}m$
In Young’s double slit experiment, angular fringe width for interference is given by,
$\beta= \dfrac {\lambda}{d}$ …(1)
Where, $\beta$ is the angular fringe width
             d is the separation between slits
             $\lambda$ is the wavelength of light
Substituting values in the equation. (1) we get,
$\beta =\dfrac { 6000\times { 10 }^{ -10 } }{ 3\times { 10 }^{ -3 } }$
$\Rightarrow \beta =2\times { 10 }^{ -4 }$
$\Rightarrow \beta=0.0002°$

So, the angular fringe width is $0.0002°$.

Note:
Angular fringe width of interference increases with decrease in distance between the slits or coherent sources and increase in wavelength of light. From the equation. (1), it can be inferred that the angular fringe width is directly proportional to the wavelength of light. Thus, the angular fringe width for red colored source or light is greater as compared to the violet colored source or violet light. Also, the angular fringe width of interference is independent of the order of fringe.