
In an interference pattern the ${\left( {n + 4} \right)^{th}}$ blue bright fringe and ${n^{th}}$ red bright fringe are formed at the spot. If ref and blue light have the wavelength of $7800\;\dot A$. and $5200\;\dot A$, the value of n should be
A. 2
B. 4
C. 6
D 8
Answer
577.8k+ views
Hint: This question is based on the interference pattern. We have to know the term interference. Interference is a phenomenon that occurs in waves. It happens when two waves coincide at the time of traveling along with the same medium. First, we have to find the expression of the centers of the bright fringes. Then equate the expression of center distance of ${\left( {n + 4} \right)^{th}}$ the blue bright fringe and ${n^{th}}$ red bright fringe to get the value of number of fringes.
Complete step by step answer:
Given: The wavelength of red light is $7800\;\dot A$ and the wavelength of blue light is $5200\;\dot A$.
The ${\left( {n + 4} \right)^{th}}$ blue bright fringe is overlapped on ${n^{th}}$ red bright fringe. So, we have to find the center distance of both the fringes.
The expression for the center distance of ${\left( {n + 4} \right)^{th}}$ the blue bright fringe is given as,
$C.D = \dfrac{{\left( {n + 4} \right)D{\lambda _B}}}{d}$
Here, $n$ is the number of fringes, \[D\] is the separation distance between the slits and the screen, and \[d\] is the distance between the slits and ${\lambda _B}$ is the wavelength of blue light.
The expression for the center distance of ${\left( {n + 4} \right)^{th}}$ the blue bright fringe is given as,
$C.D = \dfrac{{\left( n \right)D{\lambda _R}}}{d}$
Here, ${\lambda _R}$ is the wavelength of blue light.
Now, we have to equate both the expressions to get the number of fringes.
$\dfrac{{\left( {n + 4} \right)D{\lambda _B}}}{d} = \dfrac{{\left( n \right)D{\lambda _R}}}{d}$
Now, substitute the value in the above equation we get,
$\begin{array}{l}
\dfrac{{\left( {n + 4} \right)D \times 5200\;\dot A}}{d} = \dfrac{{\left( n \right)D \times 7800\;\dot A}}{d}\\
n = 8
\end{array}$
Thus, the value of $n$ is 8.
So, the correct answer is “Option D”.
Note:
In this question, students have the knowledge of the term interference and fringes. Fringes are groups that are bright or dark formed by diffraction or interference of radiation The expression of ${\left( {n + 4} \right)^{th}}$ term and ${n^{th}}$ term should be known.
Complete step by step answer:
Given: The wavelength of red light is $7800\;\dot A$ and the wavelength of blue light is $5200\;\dot A$.
The ${\left( {n + 4} \right)^{th}}$ blue bright fringe is overlapped on ${n^{th}}$ red bright fringe. So, we have to find the center distance of both the fringes.
The expression for the center distance of ${\left( {n + 4} \right)^{th}}$ the blue bright fringe is given as,
$C.D = \dfrac{{\left( {n + 4} \right)D{\lambda _B}}}{d}$
Here, $n$ is the number of fringes, \[D\] is the separation distance between the slits and the screen, and \[d\] is the distance between the slits and ${\lambda _B}$ is the wavelength of blue light.
The expression for the center distance of ${\left( {n + 4} \right)^{th}}$ the blue bright fringe is given as,
$C.D = \dfrac{{\left( n \right)D{\lambda _R}}}{d}$
Here, ${\lambda _R}$ is the wavelength of blue light.
Now, we have to equate both the expressions to get the number of fringes.
$\dfrac{{\left( {n + 4} \right)D{\lambda _B}}}{d} = \dfrac{{\left( n \right)D{\lambda _R}}}{d}$
Now, substitute the value in the above equation we get,
$\begin{array}{l}
\dfrac{{\left( {n + 4} \right)D \times 5200\;\dot A}}{d} = \dfrac{{\left( n \right)D \times 7800\;\dot A}}{d}\\
n = 8
\end{array}$
Thus, the value of $n$ is 8.
So, the correct answer is “Option D”.
Note:
In this question, students have the knowledge of the term interference and fringes. Fringes are groups that are bright or dark formed by diffraction or interference of radiation The expression of ${\left( {n + 4} \right)^{th}}$ term and ${n^{th}}$ term should be known.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: 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

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

