In a diffraction pattern due to a single slit of width 'a', the first minimum is observed at an angle 30° when light of wavelength 5000{\text \AA}  is incident on the slit. The first secondary maximum is observed at an angle of:

  • Option 1)

    {\text{sin}}^{-1}\left(\frac{1}{4}\right)

  • Option 2)

    {\text{sin}}^{-1}\left(\frac{2}{3}\right)

  • Option 3)

    {\text{sin}}^{-1}\left(\frac{1}{2}\right)

  • Option 4)

    {\text{sin}}^{-1}\left(\frac{3}{4}\right)

 

Answers (1)

As we learnt in

Fraunhofer Diffraction -

bsin 	heta = nlambda
 

- wherein

A condition of nth minima.

b= slit width

	heta = angle of deviation

 

For first minima; b\sin \Theta =\lambda

\sin \Theta =\frac{\lambda }{b}

since \Theta = 30\:\lambda =\frac{b}{2}

For 1^{st} secondary maxima

b\sin \Theta ^{1}=\frac{3\lambda }{2}

\sigma r \sin \Theta ^{1}=\frac{3\lambda }{2b}=\frac{3}{4}

\Theta ^{1}=\sin ^{-1}\left ( \frac{3}{4} \right )


Option 1)

{\text{sin}}^{-1}\left(\frac{1}{4}\right)

Incorrect

Option 2)

{\text{sin}}^{-1}\left(\frac{2}{3}\right)

Incorrect

Option 3)

{\text{sin}}^{-1}\left(\frac{1}{2}\right)

Incorrect

Option 4)

{\text{sin}}^{-1}\left(\frac{3}{4}\right)

Correct

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