The yellow light of wave length …
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Preeti Dabral 2 years, 9 months ago
Here {tex}\lambda_{1}{/tex} = 6000 {tex}\overset \circ A{/tex}, {tex}\beta_{1}{/tex} = 0.8 mm,
{tex}\lambda_{2}{/tex} = 7500 {tex}\overset \circ A{/tex}
Fringe width in first case,
{tex}\beta_{1}=\frac{D \lambda_{1}}{d}{/tex}
Fringe width in second case,
{tex}\beta_{2}=\frac{D \lambda_{2}}{2 d}{/tex}
{tex}\therefore \frac{\beta_{2}}{\beta_{1}}=\frac{D \lambda_{2} / 2 d}{D \lambda_{1} / d}=\frac{1}{2} \cdot \frac{\lambda_{2}}{\lambda_{1}}{/tex}
or {tex}\beta_{2}=\frac{1}{2} \cdot \frac{\lambda_{2}}{\lambda_{1}} \cdot \beta_{1}{/tex}
= {tex}\frac{1}{2} \times \frac{7500 \overset \circ A}{6000 \overset \circ A} \times{/tex} 0.8 mm = 0.5 mm
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