Given:
dA=1/16000 cm−1dB=1/12000 cm−1xA=3.46 cm.d_A=1/16 000 \text{ cm}^{-1}\\ d_B=1/12 000 \text{ cm}^{-1}\\ x_A=3.46\text{ cm}.dA=1/16000 cm−1dB=1/12000 cm−1xA=3.46 cm.
Solution
The fringe spacing can be expressed as
where L - distance from the screen.
Thus, for A we have
For B:
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