In a single slit diffraction experiment first minimum for red light $(660\ nm)$ coincides with first maximum of some other wavelength $\lambda$’. The value of $\lambda'$ is

Question

Dr. Abubacker · Accepted Answer

(a) In a single slit diffraction experiment, position of minima is given by $d\sin	heta = n\lambda$
So for first minima of red $\sin	heta = 1 	imes \left( \frac{\lambda_{R}}{d} ight)$
and as first maxima is midway between first and second minima, for wavelength $\lambda$’,
Its position will be
$d{\sin	heta}^{'} = \frac{\lambda^{'} + 2\lambda'}{2} \Rightarrow \sin{	heta'} = \frac{3\lambda^{'}}{2d}$
According to given condition $\sin	heta = \sin	heta'$
$\Rightarrow \lambda^{'} = \frac{2}{3}\lambda_{R}$ so $\lambda^{'} = \frac{2}{3} 	imes 660 = 440nm = 4400\mathring{\mathrm{A}}$

Q. In a single slit diffraction experiment first minimum for red light $(660\ nm)$ coincides with first maximum of some other wavelength $\lambda$’. The value of $\lambda'$ is

Explanation

Reviewed by Dr. Abubacker

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