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Interference fringes from sodium light (1=5890 \AA) in a double slit experiment have an angular width 0.20^{\circ}. To increase the fringe width by 10 \%, wavelength of light used should be

Option: 1

5892 \AA


Option: 2

4000 \AA


Option: 3

8000 \AA


Option: 4

6479 \AA


Answers (1)

best_answer

\mathrm{\beta=\frac{\lambda D}{d} \text { and angular fringe width, } \theta=\frac{\beta}{D}=\frac{\lambda}{d}}
\mathrm{\theta_{1}=\lambda_{1} / d, \theta_{2}=\lambda_{2} / d}

\mathrm{\therefore \frac{\theta_{1}}{\theta_{2}}=\frac{\lambda_{1}}{\lambda_{2}} \text { or } \lambda_{2}=\lambda_{1} \cdot \frac{\theta_{2}}{\theta_{1}}}
\mathrm{ \therefore 5890 \times \frac{0.22}{0.20}=6479 \AA}

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mansi

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