Question

What is the wavelength of light used in the experiment?

• A. $6000\mathring{A}$
• B. $5000\mathring{A}$
• C. $4800\mathring{A}$
• D. $5500\mathring{A}$

Answer 1

The correct option is A $6000\mathring{A}$

Fringe width, $\beta =\frac{\lambda D}{d}$

${\beta }_1=\frac{D_1\lambda }{d};{\beta }_2=\frac{D_2\lambda }{d}$

$\Rightarrow {\beta }_1=\frac{0.4}{100}$ and ${\beta }_2=\frac{0.4}{80}$
$\therefore {\beta }_2>{\beta }_1$ or $D_2>D_1$
$\frac{(D_2-D_1)\lambda }{d}={\beta }_2-{\beta }_1$ or

$\frac{100\lambda }{6}=0.4[\frac{1}{80}-\frac{1}{100}]$
$\frac{100\lambda }{6}=\frac{0.4\times 20}{80\times 100}$
$\lambda =\frac{8}{80\times 100}\times \frac{6}{100}cm$
$\lambda =6000\mathring{A}$