A lens of focal length f = 40 cm is cut along the diameter into two equal halves. In this process, a layer of thickness t = 1 mm is lost, then halves are put together to form a composite lens. In between focal plane and the composite lens a narrow slit is placed very close to the focal plane $|u| < |f|$ . The slit is emitting monochromatic light of wavelength $0.6 μ m$ . Behind the lens a screen is located at a distance L = 0.5 m from it as shown

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Length of interference pattern is 1/16 cm

Fringe width is 0.24 mm

Length of interference pattern is 1/8 cm

Fringe width is 0.12 mm

answer is B, C.

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Detailed Solution

$v = \frac{fu}{f - u} on the left of lens$

$d = 2 (\frac{t}{2}) (\frac{v}{u} - 1) = t (\frac{u}{f - u}) D = L + \frac{uf}{f - u}$

fringe width

$B = \frac{λ (L + \frac{uf}{f - u})}{t (\frac{u}{f - u})} = \frac{λ}{t} [f + \frac{L (f - u)}{u}] ≃ \frac{λf}{t} = \frac{6 \times 10^{- 7} \times 0.4}{10^{- 3}} = 0.24 mm$

$\begin{array}{l} \frac{y}{L - f} = \frac{t}{2 f} \Rightarrow y = \frac{t (L - f)}{2 f} l e n g t h o f i n t e r f e r e n c e p a t t e r n \\ \frac{t (L - f)}{f} + t = t (\frac{L}{f} - 1 + 1) = t \frac{L}{f} = \frac{(10 - 3) \times 50}{40} \\ = \frac{5}{4 \times 10^{3}} m = \frac{5}{40} cm = \frac{1}{8} cm \end{array}$