Question

A beam of light consisting of two wavelengths $6300\mathring{A}$ and $\lambda \mathring{A}$ is used to obtain interference fringes in a Young's double slit experiment. If $4^{th}$ bright fringe of $6300\mathring{\text{A}}$ coincides with $5^{th}$ dark fringe of $\lambda \mathring{\text{A}}$, the value of $\lambda$ is nearly $100n\mathring{\text{A}}.$ The value of $n$ is (integer only)

• A. 56.0
• B. 56
• C. 56.00

Answer 1

Answer: (A), (B), (C)

Given,

${\lambda }_1=6300\mathring{\text{A}}{\lambda }_2=\lambda \mathring{\text{A}}$

As given, 4th bright fringes of ${\lambda }_1$ coincide with $5^{th}$ dark fringe of ${\lambda }_2$

$\frac{n_1{\lambda }_{1}D}{d}=\frac{(2n_2-1){\lambda }_{2}D}{2d}$

$\left\{\begin{array}{c}\text{For dark fringe}\Delta x=\frac{(2n-1)\lambda }{2}\\ \\ \text{and for bright fringe}\Delta x=n\lambda \end{array}\right\}$

$n_1{\lambda }_1=\frac{(2n_2-1){\lambda }_2}{2}$

$4\times 6300\times {10}^{-10}=\frac{(2\times 5-1){\lambda }_2}{2}$

${\lambda }_2=5600\mathring{\text{A}}$

$\therefore n=56$