10.21 In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of n
/a.
Justify this by suitably dividing the slit to bring out the cancellation.
10.21 In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of n /a.
Justify this by suitably dividing the slit to bring out the cancellation.
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1 Answer
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10.21 Consider that a single slit of width is divided in to smaller slits.
Therefore width of each slit,
Angle of diffraction is given by the relation,
=
Now, each of these infinitesimally small slit sends zero intensity in the direction of Ø.
Hence, the combination of these slits will give zero intensity.
Similar Questions for you
At lower end
Tension, T? = 2g = 20 N (due to the 2 kg block)
Velocity, v? = √ (T? /μ) = √ (20/μ)
Wavelength, λ? = 6 cm
At upper end
Tension, T? = (2 kg + 6 kg)g = 8g = 80 N (due to the block and the rope)
Velocity, v? = √ (T? /μ) = √ (80/μ) = √4 * √ (20/μ) = 2v?
Since frequency (f) remains the same:
f = v? /λ? = v? /λ?
⇒ λ? = λ? * (v? /v? )
⇒ λ? = λ? * (2v? /v? ) = 2λ?
⇒ λ? = 2 * 6 cm = 12 cm
β = λD / (d? + a? sinωt)
β? - β? = λD/ (d? - a? ) - λD/ (d? + a? )
= λD [ (d? + a? ) - (d? - a? ) / (d? ² - a? ²) ]
= 2λDa? / (d? ² - a? ²)
3d = 0.6mm
D = 80cm
= 800mm
Path difference is given by
BP – Andhra Pradesh = Dx
[for Dark fringe at P]
n = 0, for first dark fringe
first dark fringe is observed on the screen directly opposite to one of the slits]
The distance between two successive bright fringes is fringe width .
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