Q21. The diagram below shows a section of a diffraction grating Monochromatic light of wavelength lambda is incident normally on its surface. Light waves diffracted through angle Theta form the second order image after passing through a converging lens (not shown). A, B and C are adjacent slits on the grating. (a) (i) State the phase difference between the waves at A and D. (ii)State the path length between C and E in terms of lambda Use your results to show that, for the second order image, 2lambda =dsinTheta where dis the distance between adjacent slits.

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$Q21. The diagram below shows a section of a diffraction grating Monochromatic light of wavelength lambda is incident normally on its surface. Light waves diffracted through angle Theta form the second order image after passing through a converging lens (not shown). A, B and C are adjacent slits on the grating. (a) (i) State the phase difference between the waves at A and D. __ (ii)State the path length between C and E in terms of lambda __ Use your results to show that, for the second order image, 2lambda =dsinTheta where dis the distance between adjacent slits.$

Q21. The diagram below shows a section of a diffraction grating Monochromatic light of wavelength lambda is incident normally on its surface. Light waves diffracted through angle Theta form the second order image after passing through a converging lens (not shown). A, B and C are adjacent slits on the grating. (a) (i) State the phase difference between the waves at A and D. (ii)State the path length between C and E in terms of lambda Use your results to show that, for the second order image, 2lambda =dsinTheta where dis the distance between adjacent slits.

Answer 1

(i) Phase difference between A and D is $2 \pi m$ or $m \times 360^\circ $ (ii) Path length difference between C and E is twice the distance between adjacent slits times the sine of the diffracted angle, $2d \sin \theta$ (iii) The final result showed that, for the second order image, $2 \lambda=d \sin \theta$

Answer 2

## Step1: The phase difference between waves at point A and D occurs due to the difference in the distance travelled by the wavelets from those points in order to interfere. As A and D are on adjacent slits of a diffraction grating, difference between their path length is that of the inter-slit distance d that corresponds to the order m of the spectrum. From position B, which is half-inter-slit length from both A and D, we can deduce that both wavelets are $m \lambda$ out of phase. Thus ### $\text{Phase difference between A and D} = m \times 360^\circ $, in degrees since a full wave cycle is 360 degrees, or ### $ \text{Phase difference between A and D} = 2 \pi m $, in radians where a full cycle constituting a wavelength adds 2π to phase. ## Step2: The path length difference between C and E essentially outlines a path difference for the wavelets coming through the adjacent slits much like A and D in reaching the second order image denoted by the spot E. Denoting this path difference by δ, Considering adjacent slits B (coincident with C) and D we utilize Pythagorean relationship among D-C from grating diagram, d between B & D, and δ along D-E in triangle B-D-E as follows, ### $ \delta^2 + d^2 = (d \sin \theta + d)^2 $ Solving the quadratic, for small $ \theta $, we find: ### $ \delta = 2d \sin \theta $ ## Step3: It is established that path difference is an integer multiple m of wavelength for constructive interference resulting in observable order of diffraction images. Consequently ### $ 2d \sin \theta = m \lambda $ For m=2 Substituting leads to ### $ 2 \lambda=d \sin \theta $

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