Question

In a biprism experiment, when sodium light of wavelength $5890\stackrel{\circ }{\text{A}}$ is used, then twenty fringes are observed in $23\text{mm}$ distance on screen. In order to obtain $30$ fringes in $28\text{mm}$ of the interference pattern, one should use light of wavelength

• A. $4700\stackrel{\circ }{\text{A}}$
• B. $6161\stackrel{\circ }{\text{A}}$
• C. $8835\stackrel{\circ }{\text{A}}$
• D. $4381\stackrel{\circ }{\text{A}}$

Answer 1

The correct option is A $4700\stackrel{\circ }{\text{A}}$

Given :

${\lambda }_1=5890\stackrel{\circ }{\text{A}}$

$20\text{fringes}\to 23\text{mm}$

$\because$ Distance between two fringes is given by = $\frac{\lambda D}{d}$

$\therefore$ Distance of $20$ fringes

$\frac{19{\lambda }_{1}D}{d}=23\text{mm}....\left(1\right)$

similarly distance of $30$ fringes

$\frac{29{\lambda }_{2}D}{d}=28\text{mm}...\left(2\right)$

Equation (1) / Equation (2)

$\frac{\frac{19{\lambda }_{1}D}{d}}{\frac{29{\lambda }_{2}D}{d}}=\frac{23}{28}$

$\Rightarrow \frac{19}{29}\times \frac{5890}{{\lambda }_2}=\frac{23}{28}$

${\lambda }_2=\frac{19\times 5890\times 28}{29\times 32}=4780\stackrel{\circ }{A}$