A beam of plane polarized light of large cross-sectional area and uniform intensity of falls normally on a polarizer (cross sectional area ) which rotates about its axis with an angular speed of . The energy of light passing through the polariser per revolution, is close to:
A beam of plane polarized light of large cross-sectional area and uniform intensity of falls normally on a polarizer (cross sectional area ) which rotates about its axis with an angular speed of . The energy of light passing through the polariser per revolution, is close to:
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The angle between the plane of vibration and plane of polarization is 90°.
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?
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]
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Physics NCERT Exemplar Solutions Class 12th Chapter Ten 2025
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