In young’s double slit interference experiment, the distance between two sources is 0.1/p mm. The distance of the screen from the source is 25 cm. Wavelength of light used is 5000A ?. Then what is the angular position of the first dark fringe ?”
The Correct Answer and Explanation is:
Correct Answer:
To find the angular position of the first dark fringe in Young’s double slit experiment, we use the condition for destructive interference:θdark=(2n−1)λ2d\theta_{\text{dark}} = \frac{(2n – 1)\lambda}{2d}θdark=2d(2n−1)λ
For the first dark fringe, n=1n = 1n=1, so:θ=λ2d\theta = \frac{\lambda}{2d}θ=2dλ
Given Data:
- Distance between slits d=0.1p mm=0.1×10−3p md = \frac{0.1}{p} \text{ mm} = \frac{0.1 \times 10^{-3}}{p} \text{ m}d=p0.1 mm=p0.1×10−3 m
- Wavelength λ=5000 A˚=5000×10−10 m=5×10−7 m\lambda = 5000 \text{ Å} = 5000 \times 10^{-10} \text{ m} = 5 \times 10^{-7} \text{ m}λ=5000 A˚=5000×10−10 m=5×10−7 m
Now substitute the values:θ=5×10−72×0.1×10−3p=5×10−7⋅p2×0.1×10−3=5p2×0.1×10−4=5p0.2×10−4=25p×10−4\theta = \frac{5 \times 10^{-7}}{2 \times \frac{0.1 \times 10^{-3}}{p}} = \frac{5 \times 10^{-7} \cdot p}{2 \times 0.1 \times 10^{-3}} = \frac{5p}{2 \times 0.1} \times 10^{-4} = \frac{5p}{0.2} \times 10^{-4} = 25p \times 10^{-4}θ=2×p0.1×10−35×10−7=2×0.1×10−35×10−7⋅p=2×0.15p×10−4=0.25p×10−4=25p×10−4θ=2.5p×10−3 radians\theta = 2.5p \times 10^{-3} \text{ radians}θ=2.5p×10−3 radians
So the angular position of the first dark fringe is 2.5p×10−32.5p \times 10^{-3}2.5p×10−3 radians.
Explanation
Young’s double slit experiment demonstrates the wave nature of light through the phenomenon of interference. When coherent light passes through two closely spaced slits, it creates an interference pattern of bright and dark fringes on a screen. This pattern results from the constructive and destructive interference of light waves emanating from the slits.
Constructive interference leads to bright fringes where the path difference is an integer multiple of the wavelength. Destructive interference results in dark fringes where the path difference is an odd multiple of half the wavelength. For the first dark fringe, the path difference is exactly λ2\frac{\lambda}{2}2λ.
The angular position of a dark fringe is given by:θ=(2n−1)λ2d\theta = \frac{(2n – 1)\lambda}{2d}θ=2d(2n−1)λ
Here, nnn is the order of the dark fringe, λ\lambdaλ is the wavelength of the light, and ddd is the slit separation. For the first dark fringe (n=1n = 1n=1), this simplifies to:θ=λ2d\theta = \frac{\lambda}{2d}θ=2dλ
The problem gives the slit separation as 0.1p\frac{0.1}{p}p0.1 mm, and the wavelength as 5000 angstroms, which is in the visible range. Converting all units to SI and substituting into the formula gives a general expression for the angular position: θ=2.5p×10−3\theta = 2.5p \times 10^{-3}θ=2.5p×10−3 radians. This means the angular location of the first dark fringe depends directly on the variable ppp, which would be a constant or parameter defined elsewhere.
