Question
A light source, which emits two wavelengths $\lambda_1=400 \ nm$ and $\lambda_2=600 \ nm$, is used in a Young's double slit experiment. If recorded fringe widths for $\lambda_1$ and $\lambda_2$ are $\beta_1$ and $\beta_2$ and the number of fringes for them within a distance $y$ on one side of the central maximum are $m_1$ and $m_2$, respectively, then

$(A)$ $\beta_2>\beta_1$

$(B)$ $m_1>m_2$

$(C)$ From the central maximum, $3^{\text {rd }}$ maximum of $\lambda_2$ overlaps with $5^{\text {th }}$ minimum of $\lambda_1$

$(D)$ The angular separation of fringes for $\lambda_1$ is greater than $\lambda_2$

Answer

$\beta=\frac{\lambda D}{d} \quad \therefore \quad \lambda_2>\lambda_1 \quad \text { so } \quad \beta_2>\beta_1$

No of fringes in a given width $( m )=\frac{ y }{\beta} \Rightarrow m _2< m _1$

$3^{\text {rd }}$ maximum of $\lambda_2=\frac{3 \lambda_2 D }{d}=\frac{1800 D }{ d }$

$5^{\text {th }}$ minimum of $\lambda_1=\frac{9 \lambda_1 D }{2 d }=\frac{1800 D }{ d }$

So, $3^{\text {rd }}$ maxima of $\lambda_2$ will meet with $5^{\text {th }}$ minimum of $\lambda_1$

Angular sepration $=\frac{\lambda}{ d } \Rightarrow$ Angular seperation for $\lambda_1$ will be lesser

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