Question

In a single slit diffraction experiment the first minimum for red light of wavelength $6600\mathring{A}$ coincides with the first maximum for other light of wavelength $\lambda$. The value of $\lambda$ is

• A. $2200\mathring{A}$
• B. $3300\mathring{A}$
• C. $4400\mathring{A}$
• D. $5000\mathring{A}$

Answer 1

The correct option is C $4400\mathring{A}$

Given - ${\lambda }_r=6600A^o$

The angular position of first minimum for red light of wavelength ${\lambda }_r$ , is given by

$\sin {\theta }_1=\frac{{\lambda }_r}{a}$ .......................eq1

The angular position of first maximum for other light of wavelength $\lambda$ is given by

$\sin {\theta }_1{}^{\prime }=\frac{3\lambda }{2a}$ ........................eq2

As given, the first minimum for red light coincides with the first maximum for other light, hence

$\sin {\theta }_1=\sin {\theta }_1{}^{\prime }$

or $\frac{{\lambda }_r}{a}=\frac{3\lambda }{2a}$

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

or $\lambda =\frac{2\times 6600}{3}=4400A^o$