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प्रश्न
A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young's double slit experiment on a screen placed 1 · 4 m away. If the two slits are separated by 0·28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide.
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उत्तर
Given: −
λ1 = 800 nm = 800 × 10−9 m
λ2 = 600 nm = 600 × 10−9 m
D = 1.4 m
d = 0.28 mm = 0.28 × 10−3 m
Let n1th maximum corresponds to λ1 coincides with n2th maximum corresponds to λ2. Then,
`n_1(lambda_1D)/d =n_2 ((lambda_2)D)/d`
`or,n_1/n^2 =lambda^2/lambda^1 = 600/800 =3/4`
The minimum integral value of n1 is 3 and of n2 is 4. Therefore, the minimum value of y is,
`y_(min) = n_1(lambda_1D)/d=(3 xx 800 xx 10^-9 xx 1.4)/((0.28) xx 10^-3)`
`y_(min) =12mm`
संबंधित प्रश्न
(i) In Young's double-slit experiment, deduce the condition for (a) constructive and (b) destructive interferences at a point on the screen. Draw a graph showing variation of intensity in the interference pattern against position 'x' on the screen.
(b) Compare the interference pattern observed in Young's double-slit experiment with single-slit diffraction pattern, pointing out three distinguishing features.
In Young's double slit experiment, plot a graph showing the variation of fringe width versus the distance of the screen from the plane of the slits keeping other parameters same. What information can one obtain from the slope of the curve?
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The separation between the consecutive dark fringes in a Young's double slit experiment is 1.0 mm. The screen is placed at a distance of 2.5m from the slits and the separation between the slits is 1.0 mm. Calculate the wavelength of light used for the experiment.
In a Young's double slit interference experiment, the fringe pattern is observed on a screen placed at a distance D from the slits. The slits are separated by a distance d and are illuminated by monochromatic light of wavelength \[\lambda.\] Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one-fourth the maximum.
Draw a neat labelled diagram of Young’s Double Slit experiment. Show that `beta = (lambdaD)/d` , where the terms have their usual meanings (either for bright or dark fringe).
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Justify the above statement through a relevant mathematical expression.
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In Young's double slit experiment, the distance of the 4th bright fringe from the centre of the interference pattern is 1.5 mm. The distance between the slits and the screen is 1.5 m, and the wavelength of light used is 500 nm. Calculate the distance between the two slits.
