Q143 / 180 · KCET 2018 · Physics
In Young's double slit experiment, two wavelengths \( \lambda_{1}=780 \mathrm{~nm} \) and \( \lambda_{2}=520 \mathrm{~nm} \) are used to
obtain interferance fringes. If the \( n^{\text {th }} \) bright band due to \( \lambda_{1} \) concides with \( (n+1)^{\text {th }} \) bright band
due to \( \lambda_{2} \) then the value of \( n \) is
obtain interferance fringes. If the \( n^{\text {th }} \) bright band due to \( \lambda_{1} \) concides with \( (n+1)^{\text {th }} \) bright band
due to \( \lambda_{2} \) then the value of \( n \) is
- A\( 04 \)
- B\( 03 \)
- Correct answerC\( 12 \)
- D\( 06 \)
Solution
Given,
\( n^{\text {th }} \) bright band due to \( \lambda_{1}=(n+1)^{\text {th }} \) bright band due to \( \lambda_{2} \)
\[ \begin{array}{l} \Rightarrow \frac{n \lambda_{1} D}{d}=\frac{(n+1) \lambda_{2} D}{d} \\ \Rightarrow n \lambda_{1}=(n+1) \lambda_{2} \\ \Rightarrow \frac{n+1}{n}=\frac{\lambda_{1}}{\lambda_{2}} \end{array} \]
Given, \( \lambda_{1}=780 \mathrm{~nm} \) and \( \lambda_{2}=520 \mathrm{~nm} \), therefore
\[ \frac{n+1}{n}=\frac{780}{520} \]
On rearranging, we get
\[ \begin{array}{l} 52(n+1)=78 n \\ \Rightarrow 52 n+52=78 n \\ \Rightarrow 78 n-52 n=52 \\ \Rightarrow 26 n=52 \\ n=\frac{52}{26}=2 \end{array} \]
Thus, value of \( n \) is \( 2 \).
\( n^{\text {th }} \) bright band due to \( \lambda_{1}=(n+1)^{\text {th }} \) bright band due to \( \lambda_{2} \)
\[ \begin{array}{l} \Rightarrow \frac{n \lambda_{1} D}{d}=\frac{(n+1) \lambda_{2} D}{d} \\ \Rightarrow n \lambda_{1}=(n+1) \lambda_{2} \\ \Rightarrow \frac{n+1}{n}=\frac{\lambda_{1}}{\lambda_{2}} \end{array} \]
Given, \( \lambda_{1}=780 \mathrm{~nm} \) and \( \lambda_{2}=520 \mathrm{~nm} \), therefore
\[ \frac{n+1}{n}=\frac{780}{520} \]
On rearranging, we get
\[ \begin{array}{l} 52(n+1)=78 n \\ \Rightarrow 52 n+52=78 n \\ \Rightarrow 78 n-52 n=52 \\ \Rightarrow 26 n=52 \\ n=\frac{52}{26}=2 \end{array} \]
Thus, value of \( n \) is \( 2 \).
Hence, the correct answer is option: 'C'.