A lens of focal length 'f ' is cut along the diameter into two identical halves. In this process a layer of the lens 't' in thickness is lost, then the halves are put together to form a composite lens. In between the focal plane and the composite lens, a narrow slit is placed very near the focal plane. The slit is emitting monochromatic light with wavelength λ . Behind the lens a screen is located at a distance L from it –
(i)Fringe width of Interference pattern –
Text Solution
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Ans.
(i)
Sol.
= + 
v = 
as u < f ∴ v = – ve
| v | = 
m = 
I = 
d = 2I – t = 
– t = 
D = L + | v |
= L + 
w = fringe width = 
= 
= 
as u → f
w = 
(ii)
Sol. 
= 
x = 
= 
= 
length of interference zone = 2x = 
(iii) [ a]
Sol. Number of maxima
= 
× 2 = 
× 2
= 
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