In a YDSE both slits produce equal intensities on the screen. A 100 % transparent thin film is placed in front of one of the slits. Now the intensity of the geometrical centre of system on the screen becomes 75 % of the previous intensity. The wavelength of the light is 6000Å and μ glass = 1.5. The thickness of the film cannot be:
Text Solution
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As we know, at the point of 75% intensity
cos φ = 
{Refer Q B–4, Exercise – I Objective}
⇒ 
× Δ P = 
, 
, 
, 
, 
⇒ (µ – 1) t = 
⇒ t = 
= 0.2 μ m; 1 μ m, 1.4 μ m, 2.6 μ m.......
Hence only 1.6 μ m is not possible.
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