Question

The fringe width of interference pattern obtained at screen is

• A. $1.5\times {10}^{-4}m$
• B. $1.7\times {10}^{-4}m$
• C. $1.7\times {10}^{-5}m$
• D. $1.5\times {10}^{-5}m$

Answer 1

The correct option is B $1.7\times {10}^{-4}m$

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

$\frac{1}{20}+\frac{1}{v}=\frac{1}{5}$

$v=\frac{20}{3}cm$

$m=\frac{-v}{u}=-\frac{1}{3}$

$m=\frac{{-h}_i}{h_o}$

$h_0=1/2,h_i=1/6$

$d=2\left(\frac{1}{6}\right)+1$

$d=\frac{4}{3}mm\Rightarrow D=\frac{340}{3}cm$

$\beta =\frac{\lambda D}{d}=\frac{\left(\lambda \right)3400}{4}=\left(\lambda \right)850$

$\beta =2000\times 850\times {10}^{-10}\Rightarrow \beta =1.7\times {10}^{-4}m$