Consider sunlight incident on a pinhole of width 103 Å. The image of the pinhole seen on a screen shall be
(a) A sharp white ring
(b) Different from a geometrical image
(c) A diffused central spot, white in colour
(d) Diffused coloured region around a sharp central white spot
Consider sunlight incident on a pinhole of width 103 Å. The image of the pinhole seen on a screen shall be
(a) A sharp white ring
(b) Different from a geometrical image
(c) A diffused central spot, white in colour
(d) Diffused coloured region around a sharp central white spot
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1 Answer
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This is a Multiple Choice Questions as classified in NCERT Exemplar
Answer – (b, d)
Explanation- We know that wavelength of sunlight ranges from 4000 Å to 8000 Å.
Clearly, wavelength λ < width of the slit.
Hence, light is diffracted from the hole. Due to diffraction from the slight the image formed On the screen will be different from the geometrical image.
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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