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In a Young’S Double-slit Experiment, the Slits Are Separated by 0.28 Mm and the Screen is Placed 1.4 M Away. the Distance Between the Central Bright Fringe and the Fourth Bright Fringe is

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प्रश्न

In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

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उत्तर

Distance between the slits, d = 0.28 mm = 0.28 × 10−3 m

Distance between the slits and the screen, D = 1.4 m

Distance between the central fringe and the fourth (n = 4) fringe, u = 1.2 cm = 1.2 × 10−2 m

In case of constructive interference, we have the relation for the distance between the two fringes as:

u = `nlambda D/d`

Where,

n = Order of fringes = 4

λ = Wavelength of light used

∴ `lambda = (ud)/(nD)`

= `(1.2 xx 10^(-2) xx 0.28 xx 10^(-3))/(4xx1.4)`

= 6 × 10−7

= 600 nm

Hence, the wavelength of the light is 600 nm.

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पाठ 10: Wave Optics - EXERCISES [पृष्ठ २७३]

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एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 10 Wave Optics
EXERCISES | Q 10.4 | पृष्ठ २७३
एनसीईआरटी Physics Part I and II [English] Class 12
पाठ 10 Wave Optics
Exercise | Q 10.4 | पृष्ठ ३८३

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