Question

Bichromatic light of wavelengths ${\lambda }_1=5000\dot{\text{A}}\text{and}{\lambda }_2=7000\dot{\text{A}}$ are used in $YDSE$. Then,

• A. ${14}^{th}$ order maxima of ${\lambda }_1$, will coincide with ${10}^{th}$ order maxima of ${\lambda }_2$.
  
• B. ${21}^{st}$ order maxima of ${\lambda }_2$, will coincide with ${15}^{th}$ order maxima of ${\lambda }_1$.
• C. ${11}^{th}$ order minima of ${\lambda }_1$, will coincide with $8^{th}$ order minima of ${\lambda }_2$.
• D. $3^{rd}$ order minima of ${\lambda }_1$, will coincide with $4^{th}$ order maxima of ${\lambda }_2$

Answer 1

Answer: (A), (C)

${11}^{th}$ order minima of ${\lambda }_1$, will coincide with $8^{th}$ order minima of ${\lambda }_2$.

Condition for overlapping of maxima,

$\frac{n_1{\lambda }_{1}D}{d}=\frac{n_2{\lambda }_{2}D}{d}$

$\Rightarrow \frac{n_1}{n_2}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{7}{5},\frac{14}{10}.......=\frac{7x}{5x}$

For, $x=2\to n_1=14\text{and}{n}_2=10$

$\Rightarrow$${14}^{th}$ order maxima of ${\lambda }_1$, will coincide with ${10}^{th}$ order maxima of ${\lambda }_2$.

Condition for overlapping of minima,

$\frac{(2n_1-1){\lambda }_{1}D}{2d}=\frac{(2n_1-1){\lambda }_{1}D}{2d}$

$\Rightarrow \frac{2n_1-1}{2n_2-1}=\frac{{\lambda }_2}{{\lambda }_1}=\frac{7}{5},\frac{14}{10}.....=\frac{7y}{5y}$

For $y=3\Rightarrow n_1=11\text{and}{n}_2=8$

$\Rightarrow$${11}^{th}$ order minima of ${\lambda }_1$, will coincide with $8^{th}$ order minima of ${\lambda }_2$.

Hence, option $\left(A\right)$ and $\left(C\right)$ is correct.