Wave Optics — CBSE Class 12 · 2017 · Delhi Board Question with Solution

(a) Explain two features to distinguish between the interference pattern in Young's double slit experiment and the diffraction pattern obtained due to a single slit.
(b) A monochromatic light of wavelength

500

nm is incident normally on a single slit of width

0.2

mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen. Estimate the number of fringes obtained in Young's double slit experiment with fringe width

0.5

mm, which can be accommodated within the region of total angular spread of the central maximum due to the single slit.

Solution

(a) Two distinguishing features:
(i) In interference all bright fringes are equally bright and equally spaced; in diffraction the central maximum is by far the brightest and the side bands fall off in intensity.
(ii) Interference results from two separate coherent wavefronts (two slits); diffraction results from secondary wavelets of a single wavefront (one slit), and its central band is twice as wide as the side bands.
(b) Angular width of the central maximum:

2 θ = \frac{2 λ}{a} = \frac{2 \times 500 \times 1 0 ^{- 9}}{0.2 \times 1 0 ^{- 3}} = 5 \times 1 0^{- 3} rad

Linear width on a screen at distance

D

ω^{'} = \frac{2 λ D}{a}

. The number of YDSE fringes of width

β

that fit in this spread is

n = \frac{ω ^{'}}{β} = \frac{2 λ D / a}{β}

Taking the linear central width

= 2 θ \cdot D

and

β = 0.5

mm with

D = 1

ω^{'} = 5 \times 1 0^{- 3} \times 1 = 5 \times 1 0^{- 3}

m, so

n = \frac{5 \times 1 0 ^{- 3}}{0.5 \times 1 0 ^{- 3}} = 10

Using

D

such that the full angular spread corresponds to

ω^{'} = 10

mm gives about

20

fringes; with

D = 1

m as above,

n \approx 10

fringes are accommodated.

2 θ = 5 \times 1 0^{- 3} rad; n \approx 10 - 20 fringes

CBSE Class 12 Board · 2017 · Delhi · Subjective

CBSE Class 12 · 2017 · Delhi · Wave OpticsLong Answer

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