wave-optics Question 23

Question: Q. 6. Consider a two slit interference arrangement (shown in figure) such that the distance of the screen from the slits is half the distance between the slits. Obtain the value of D in terms of λ such that the first minima on the screen fall at a distance D from the centre O.

U] [SQP 2016]

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Solution:

Ans.

Minima will occur when

=[D2+(D+x)2]1/2 [D2+(Dx)2]1/2 =λ2

If

(D2+4D2)1/2(D2)1/2=λ2

(5D2)1/2(D2)1/2=λ2

D(51)=λ2

D=λ2(51)

[CBSE Marking Scheme 2016]

[II Q. 7. Explain by drawing a suitable diagram that the interference pattern in a double slit is actually a superposition of single slit diffraction from each slit. Write two basic features which distinguish the interference pattern from those seen in a coherently illuminated single slit. U [Delhi 2015]

Ans. The diagram, given here, shows several fringes, due to double slit interference, ‘contained’ in a broad diffraction peak. When the separation between the slits is large compared to their width, the diffraction pattern becomes very flat and we observe the two slit interference pattern.

1

Two basic features :

(i) The interference pattern has a number of equally spaced bright and dark bands while diffraction pattern has a central bright maxima which is twice as wide as the other maxima.

(ii) Interference pattern is the superposition of two waves originating from two narrow slits. The diffraction pattern is a superposition of a continuous family of waves originating from each point on a single slit.

(iii) For a single slit of width ’ a ’ the first null of

2 diffraction pattern occurs at an angle of λa. At the same angle of λ/a, we get a maxima for two narrow slits separated by a distance a.

(Any two) 1



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