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
543.5k+ views
- 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 $\dfrac{\pi }{{{180}^{{}^\circ }}}$.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Discuss the various forms of bacteria class 11 biology CBSE

