A lens of focal length $f = 40 c m$ is cut along the diameter in two identical halves. In this process, a layer of the lens $t = 1 m m$ 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, near the focal plane. The slit is emitting monochromatic light with wavelength $λ = 0.6 μ m$ . Behind the lens a screen is located at a distance $L = 0.5 m$ from it. Find the fringe width and number of visible maxima.

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$\frac{L t^{2}}{2 λ f^{2}}$

$\frac{L t}{2 λ f}$

$- \frac{L t}{2 λ f}$

$- \frac{L t^{2}}{2 λ f^{2}}$

answer is A.

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Detailed Solution

From lens equation, we have

$v = \frac{u f}{f - u} (v lines on the left side)$

$d = 2 (\frac{t}{2}) [\frac{u}{v} - 1] = t [\frac{u}{f - u}]$

$D = L + v = L + \frac{u f}{f - u}$

Fringe width,

$ω = \frac{λ D}{d} = \frac{λ [L + \frac{u}{f - u}]}{t [\frac{u}{f - u}]}; or ω = \frac{λ}{t} [f + \frac{L (f - u)}{u}]$

as $u \to f . ω = \frac{λ f}{t}$

From figure, $\frac{y}{L - f} = \frac{t / 2}{f} \Rightarrow y = \frac{t / 2}{f} (L - f)$

Distance of the point up to which interference occurs,

$y_{0} = \frac{t / 2}{f} (L - f) + \frac{t}{2} = \frac{t L}{2 f}$

Number of visible maxima $= \frac{t L . t}{2 f \times λ f} = \frac{L t^{2}}{2 λ f^{2}}$

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