In an interference experiment as shown in the figure, the source plane and screen are separated by a distance 1m. At a certain position of source, fringe width is $\frac{1}{4}$ mm and by moving the source a way from mirror along the line AB by 0.6 mm, the fringe width changed to $\frac{1}{6} m m$ :-

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Wavelength of light used is $5000 \overset{0}{A}$

Initial distance of source from A is 0.6 mm

Wavelength of light used is $6000 \overset{0}{A}$

Initial distance of source from A is 1.2 mm

answer is B, C.

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Detailed Solution

$AS = h \frac{1}{4} = \frac{λ (1)}{2 h} . . . . . . (i)$

$\begin{array}{ll} \frac{1}{6} = \frac{λ (1)}{2 (h + 0.6)} \dots \dots \dots \dots . (ii) \\ \frac{\frac{1}{4}}{\frac{1}{6}} = \frac{h + 0.6}{h} \frac{3}{2} = \frac{h + 0.6}{h} \end{array}$

$\begin{array}{l} h = 1.2 mm \frac{1}{4} m m = \frac{λ (1 m)}{2 \times (1.2) mm} \\ λ = 0.6 \times 10^{- 6} m λ = 6000 A \end{array}$

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