Question

A single slit diffraction pattern is formed with white light. For what wavelength of light, the third secondary maximum in the diffraction pattern coincides with the second secondary maximum of red light in the pattern. Take wavelength of red light as $6500\mathring{\text{A}}$?

• A. $6420\mathring{\text{A}}$
• B. $4120\mathring{\text{A}}$
• C. $4640\mathring{\text{A}}$
• D. $8840\mathring{\text{A}}$

Answer 1

The correct option is C $4640\mathring{\text{A}}$

For a secondary maxima, $a\sin \theta =\frac{(2m+1)\lambda }{2}$
For 2nd secondary maxima using red light,
$a\sin {\theta }_1=\frac{5{\lambda }_1}{2}\Rightarrow \sin {\theta }_1=\frac{5{\lambda }_1}{2a}...\left(i\right)$

For third secondary maxima using the unknown wavelength,
$a\sin {\theta }_2=\frac{7{\lambda }_2}{2}\Rightarrow \sin {\theta }_2=\frac{7{\lambda }_2}{2}...\left(ii\right)$

Since, these positions coincide with each other for unknown wavelength and red light,
$\Rightarrow \sin {\theta }_1=\sin {\theta }_2$

$\Rightarrow \frac{5{\lambda }_1}{2a}=\frac{7{\lambda }_2}{2a}$
$\Rightarrow {\lambda }_2=\frac{5}{7}{\lambda }_1=\frac{5}{7}\times 6500$
$\therefore {\lambda }_2\approx 4640\stackrel{\circ }{A}$

Hence, $\left(C\right)$ is the correct answer.