Question

In a single slit diffraction experiment, first minimum for red light $\left(660\text{nm}\right)$ coincides with first maximum of some other light of wavelength $\lambda$. The value of $\lambda$ in $\text{nm}$ is

• A. $440\text{nm}$
• B. $540\text{nm}$
• C. $470\text{nm}$
• D. $510\text{nm}$

Answer 1

The correct option is A $440\text{nm}$

Wavelength of red light, ${\lambda }_r=660\text{nm}$

The condition for minima is given by,

$\Rightarrow \sin \theta =\frac{n\lambda }{b}$

$\therefore$ For first minima $(n=1)$ of red light,

$\sin \theta =\frac{1\times {\lambda }_r}{b}$

The condition for maxima is given by,

$b\sin \theta =(2n+1)\frac{\lambda }{2}$

$\therefore$ For first maxima $(n=1)$ of $\lambda$

$\sin {\theta }^{\prime }=\frac{3\lambda }{2b}$

As, the two coincide, angular positions will be same

$\Rightarrow \sin \theta =\sin {\theta }^{\prime }$

$\Rightarrow \frac{{\lambda }_r}{b}=\frac{3\lambda }{2b}$

$\Rightarrow \lambda =\frac{2}{3}{\lambda }_r$

$\Rightarrow \lambda =\frac{2}{3}\times 660\text{nm}=440\text{nm}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.