Question: Q. 5. Laser light of wavelength $640 \mathrm{~nm}$ incident on a pair of slits produces an interference pattern in which the bright fringes are separated by $7.2 \mathrm{~mm}$. Calculate the wavelength of another source of light which produces interference fringes separated by $8.1 \mathrm{~mm}$ using same arrangement. Also find the minimum value of the order $(n)$ of bright fringe of shorter wavelength which coincides with that of the longer wavelength. A [O.D. Comptt. I, II, III 2012]
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Solution:
Ans. Fringe width in interference pattern
$$ \begin{equation*} \beta=\frac{D \lambda}{d} \tag{i} \end{equation*} $$
According to the question when $\lambda=640 \mathrm{~nm}$ then $\beta=7.2 \mathrm{~mm}$. Putting these values in the relation (i)
$$ \begin{aligned} 7.2 \times 10^{-3} & =\frac{D \times 640 \times 10^{-9}}{d} \ \frac{D}{d} & =\frac{7.2 \times 10^{-3}}{640 \times 10^{-9}}=\frac{7.2}{640 \times 10^{-6}} \end{aligned} $$
Now with another monochromatic light $\lambda^{\prime}$, the fringe width $\beta^{\prime}=8.1 \mathrm{~mm}$. Hence
$$ 8.1 \times 10^{-3}=\frac{D \times \lambda^{\prime}}{d} $$
Putting the value of $\frac{D}{d}$
$$ \begin{aligned} 8.1 \times 10^{-3} & =\frac{7.2}{640 \times 10^{-6}} \times \lambda^{\prime} \ \lambda^{\prime} & =\frac{640 \times 8.1 \times 10^{-9}}{7.2} \ & =720 \mathrm{~nm} \end{aligned} $$
Condition for minimum value of the order $(n)$ of bright fringe of shorter wavelength which coincides with that of the longer wavelength
$$ \begin{aligned} \frac{n_{1} D}{d} \lambda & =\frac{n_{2} D}{d} \lambda^{\prime} \ \frac{n_{1}}{n_{2}} & =\frac{\lambda^{\prime}}{\lambda} \ \frac{n_{1}}{n_{2}} & =\frac{720 \times 10^{-9}}{640 \times 10^{-9}} \ \frac{n_{1}}{n_{2}} & =\frac{9}{8} \end{aligned} $$
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