Question

Monochromatic light of wavelengths $400$ nm and $560$ nm are incident simultaneously and normally on double-slit apparatus whose slit separation is $0.1$ mm and screen distance is $1m$. Distance between areas of total darkness will be ___ mm

• A. $4$
• B. $5.6$
• C. $14$
• D. $28$

Answer 1

The correct option is D $28$

Given:

Wavelengths, ${\lambda }_1=400nm=400\times {10}^{-9}m$,

${\lambda }_2=560nm=560\times {10}^{-9}m$

Slit separation, $d=0.1mm=0.1\times {10}^{-3}m$

Screen distance, $D=1m$

To find:

Distance between areas of total darkness, $\Delta s=?$

According to the question,

$(2n+1){\lambda }_1=(2m+1){\lambda }_2⟹\frac{(2n+1)}{(2m+1)}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{560nm}{400nm}=\frac{7}{5}$

$10n+5=14m+7⟹10n=14m+2⟹5n=7m+1$

By inspection, for $m_1=2,n_1=3$ and $m_2=7,n_2=10$

Therefore, $\Delta s=\frac{{\lambda }_{1}D}{2d}\left[\right(2n_2+1)-(2n+1\left)\right]$

$=\frac{400\times {10}^{-9}\times 1}{2\times 0.1\times {10}^{-3}}\left[\right(2\times 10+1)-(2\times 3+1\left)\right]$

$=200\times {10}^{-5}(21-7)$

$⟹\Delta s=2800\times {10}^{-5}=28mm$

Hence, the distance between areas of total darkness will be $28mm$