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Interference of Light

Problem 5.73

A lens of diameter 5.0 cm5.0 \mathrm{~cm} and focal length f=25.0 cmf=25.0 \mathrm{~cm} was cut along the diameter into two identical halves. In the process, the layer of the lens a=1.00 mma=1.00 \mathrm{~mm} in thickness was lost. Then the halves were put together to form a composite lens. In this focal plane a narrow slit was placed, emitting monochromatic light with wavelength λ=0.60μm.\lambda=0.60 \mu \mathrm{m} . Behind the lens a screen was located at aa distance b=50 cmb=50 \mathrm{~cm} from it. Find: (a) the width of a fringe on the screen and the number of possible maxima; (b) the maximum width of the slit δmax⁡\delta_{\max } at which the fringes on the screen will be still observed sufficiently sharp.

Reveal Answer
5.73. (a) Δx=λf/a=0.15 mm,13\Delta x=\lambda f / a=0.15 \mathrm{~mm}, 13 maxima; (b) the fringes are still sufficiently sharp when δx⩽Δx/2,\delta x \leqslant \Delta x / 2, where δx\delta x is the shift of the fringes from the extreme elements of the slit, hence, δmax⁡=\delta_{\max }= =λf2/2ab=37μm=\lambda f^{2} / 2 a b=37 \mu \mathrm{m}
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