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  • In a single slit diffraction experiment, first minimum for red light (660 nm) coincides with first maximum of some other wavelength <img alt="\lambda^{\prime}" src="https://entrancecorner.oncodecog

In a single slit diffraction experiment, first minimum for red light (660 nm) coincides with first maximum of some other wavelength \lambda^{\prime} The value of \lambda^{\prime} is:

Option: 1

440 \mathrm{~nm}


Option: 2

550 \mathrm{~nm}


Option: 3

600 \mathrm{~nm}


Option: 4

650 \mathrm{~nm}


Answers (1)

best_answer

Here, \lambda_{r}=660 \mathrm{~nm} ; \quad \lambda^{\prime}= ?

For diffraction minima,
a \sin \theta=n \lambda, \sin \theta=\frac{n \lambda}{a}
For first minima of red,
\sin \theta=\frac{(1) \lambda_{r}}{a}
For diffraction maxima
a \sin \theta=(2 n+1) \frac{\lambda}{2}
\therefore  For first maxima of \lambda^{\prime}
a \sin \theta^{\prime}=\frac{3 \lambda^{\prime}}{2} ; \quad \sin \theta^{\prime}=\frac{3 \lambda^{\prime}}{2 a}

As the two coincide, therefore,
\sin \theta^{\prime}=\sin \theta
\frac{3 \lambda^{\prime}}{2 \mathrm{a}}=\frac{\lambda_{r}}{\mathrm{a}}
or \quad \lambda^{\prime}=\frac{2}{3} \lambda_{r}
or \quad \lambda^{\prime}=\frac{2}{3}(660)=440 \mathrm{~nm}

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manish

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