
Find the angular fringe width in young’s double-slit experiment with a blue-green light of wavelength 6000\[{{A}^{{}^\circ }}\]. The separation between the slits is $3.0\times {{10}^{-3}}m$.
Answer
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- Hint: This is the question from optics. The angular fringe width is the distance of the central fringe from the slit. It also indicates the angular separation between the fringes. Since wavelength and separation between slits are given, use the formula of angular fringe width in Young's double-slit experiment which will give the relation between wavelength and separation between two slits.
Complete step-by-step solution:
In young double slits experiment angular fringe width, also known as bandwidth is given by,
$\theta =\dfrac{\lambda }{d}$
Where,
\[\lambda \]= the wavelength of blue-green light
d= the distance between two adjacent slits or the separation between slits.
We have given the wavelength of the light and the separation between slits.
$\begin{align}
& \lambda =6000{{A}^{0}}=6000\times {{10}^{-10}}m \\
& d=3\times {{10}^{-3}}m \\
\end{align}$
Therefore,
Angular fringe width is given as
$\begin{align}
& \theta =\dfrac{6000\times {{10}^{-10}}}{3\times {{10}^{-3}}} \\
& \theta =2\times {{10}^{-4}} \\
& \theta ={{0.0002}^{0}} \\
\end{align}$
The angular fringe width in young’s double slit experiment with a blue-green light of wavelength 6000\[{{A}^{{}^\circ }}\] is $\theta ={{0.0002}^{0}}$.
Additional information
Young’s double slit was the first experiment to demonstrate (observe) interference of light.
Using this experiment, Young’s proved that light is propagated in the form of waves.
The wavelength of monochromatic light can be determined $\lambda =\dfrac{Xd}{D}$
Note: In Young’s double-slit experiment a monochromatic source of light is illuminated on closely spaced narrow slits. The fringe pattern depends on the wavelength of light used. The fringe width depends on the wavelength of light used, the distance between two slits, and the distance of the screen from slits. From the formula, you can note that angular fringe width is directly proportional to the wavelength of light used and inversely proportional to the distance between two slits. Therefore, the angular fringe width will increase with an increase in wavelength and will decrease with an increase in distance between two slits. Also note that the final answer is in degrees not in radian, if you want the answer in radian then multiply the answer in degree by $\dfrac{\pi }{{{180}^{{}^\circ }}}$.
Complete step-by-step solution:
In young double slits experiment angular fringe width, also known as bandwidth is given by,
$\theta =\dfrac{\lambda }{d}$
Where,
\[\lambda \]= the wavelength of blue-green light
d= the distance between two adjacent slits or the separation between slits.
We have given the wavelength of the light and the separation between slits.
$\begin{align}
& \lambda =6000{{A}^{0}}=6000\times {{10}^{-10}}m \\
& d=3\times {{10}^{-3}}m \\
\end{align}$
Therefore,
Angular fringe width is given as
$\begin{align}
& \theta =\dfrac{6000\times {{10}^{-10}}}{3\times {{10}^{-3}}} \\
& \theta =2\times {{10}^{-4}} \\
& \theta ={{0.0002}^{0}} \\
\end{align}$
The angular fringe width in young’s double slit experiment with a blue-green light of wavelength 6000\[{{A}^{{}^\circ }}\] is $\theta ={{0.0002}^{0}}$.
Additional information
Young’s double slit was the first experiment to demonstrate (observe) interference of light.
Using this experiment, Young’s proved that light is propagated in the form of waves.
The wavelength of monochromatic light can be determined $\lambda =\dfrac{Xd}{D}$
Note: In Young’s double-slit experiment a monochromatic source of light is illuminated on closely spaced narrow slits. The fringe pattern depends on the wavelength of light used. The fringe width depends on the wavelength of light used, the distance between two slits, and the distance of the screen from slits. From the formula, you can note that angular fringe width is directly proportional to the wavelength of light used and inversely proportional to the distance between two slits. Therefore, the angular fringe width will increase with an increase in wavelength and will decrease with an increase in distance between two slits. Also note that the final answer is in degrees not in radian, if you want the answer in radian then multiply the answer in degree by $\dfrac{\pi }{{{180}^{{}^\circ }}}$.
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