Q89 / 180 · AIIMS 2008 · Physics

Light is incident normally on a diffraction grating through which the first diffraction is seen at \(32^{\circ}\). In this case the second order diffraction will be

  • A
    at \(80^{\circ}\)
  • B
    at \(64^{\circ}\)
  • C
    at \(48^{\circ}\)
  • Correct answer
    D
    there is no second order diffraction

Solution

Angle of first order diffraction $\left(\theta_1\right)=32^{\circ}$. We know that the angle of diffraction for the $n$th order $\left(\theta_n\right)$ is given by $d \sin \theta_n=n \lambda$. For first order diffraction we get $d \sin 32^{\circ}=1 \times \lambda$ or, $\lambda=d \sin 32^{\circ}$. Now for second order diffraction, $d \sin \theta_2=2 \times \lambda$ or, $d \sin \theta_2=2 \times d \sin 32^{\circ}$ or $\sin \theta_2=2 \times \sin 32^{\circ}=2 \times 0.529=1.06$. Since the sine of any angle cannot be greater than 1 , therefore there is no second order diffraction.

Hence, the correct answer is option: 'D'.

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