Question

Originally the Young's double slit experiment was performed in white light in which higher order spectra in the fringe pattern overlap each other. Assume that wavelength of blue light is 520 nm and that of red light is 780 nm. The value of x for which $x^{th}$ bright red band coincides with (x+1) th blue band is

• A. 5
• B. 4
• C. 3
• D. 2

Answer 1

The correct option is D 2

$\begin{array}{c}\text{Given that,}\\ \text{-}\lambda \text{blue}=520nm\lambda =\left(\text{wavelength of light}\right)\\ \text{-}\lambda \text{Red}=780nm\\ x^{\text{th}}\text{bright red band coincide with}(x+1{)}^{\text{th}}\text{blue band}\end{array}$

$\begin{array}{c}\text{it means distance from centre forthred band is equal}\\ (x+1{)}^{\text{th}}\text{bue band.}\\ \text{we know,}\\ \text{Distance from centre for ony bright band}=\frac{n\lambda D}{d}\end{array}$

$\text{For Red}$

$d_{\gamma }=\frac{x{\lambda }_{r}D}{d}$

$\text{For blue}$

$\begin{array}{cc}{d}_b& =\frac{(x+1){\lambda }_{b}D}{d}\\ d_r& =d_b\left(\text{given}\right)\end{array}$

$\text{Solving (i) and (ii)}$

$\left(\frac{x+1}{x}\right)=\frac{{\lambda }_r}{x_b}$

$\begin{array}{cc}\frac{x+1}{x}& =\frac{780}{520}\\ \frac{x+1}{x}& =\frac{6}{4}\\ 4x+4& =6x\\ 2x=4& \\ x=2& \end{array}$

$\therefore$

$2^{\text{nd}}\text{band of red coincide with}$

$3^{\text{rd}}\text{band of blue}$