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# In a biprism experiment, the micrometer readings for the zero-order and ${10^{th}}$ order fringe are $1.25mm$ and $2.37mm$ respectively, when the light of $600nm$ is used. If the wavelength is changed to $750nm$, What will be the respective position of zero and ${10^{th}}$ order fringes?A. $1.25mm,2.65mm$B. $1.52mm,2.56mm$C. $1.26mm,2.28mm$D. $1.32mm,2.65mm$  Verified
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Hint: Calculate the zero error of micrometer as the reading of micrometer for zero order is non-zero.
The position of ${n^{th}}$ order fringe is given by ${y_n} = \dfrac{{n\lambda D}}{d}$ where $n$ is the order of the fringe, $\lambda$ is the wavelength of the light used, $D$ is the distance between the source and the screen and $d$ is the distance between the slits.

Complete step by step solution:
It is given in the question that the micrometer readings for the zero-order and ${10^{th}}$ order fringe are $1.25mm$ and $2.37mm$ respectively when the light of $600nm$ is used. Since the reading of the micrometer for zero-order is non-zero, there is a zero error which is equal to $1.25mm$.
Hence this zero error needs to be removed to find the true positions of fringes.
Since, the distance between zero order and ${10^{th}}$ order is $2.37mm - 1.25mm = 1.12mm$ .
Now, the position of ${n^{th}}$ order fringe is given by ${y_n} = \dfrac{{n\lambda D}}{d}$ where $n$ is the order of the fringe, $\lambda$ is the wavelength of the light used, $D$ is the distance between the source and the screen and $d$ is the distance between the slits.
As, $D$ and $d$ are constant for this experiment then we can say that
$\dfrac{{\Delta {y_n}}}{{{y_n}}} = \dfrac{{\Delta \lambda }}{\lambda }$
Where, $\Delta \lambda = 750nm - 600nm = 150nm$ and for $\lambda = 600nm$, ${y_n} = 2.37mm$ for ${10^{th}}$ order fringe.
So, $\dfrac{{\Delta {y_n}}}{{{y_n}}} = \dfrac{{150}}{{600}} = 0.25$ or we can say that $\Delta {y_n} = 0.25{y_n}$
Now, $y_n^{'} = {y_n} + \Delta {y_n} = {y_n} + 0.25{y_n}$
So, $y_n^{'} = 1.25{y_n} = 1.25 \times 1.12mm = 1.4mm$ (as, ${y_n} = 1.12mm$)
Adding the zero error of the micrometer gives the new position of ${10^{th}}$ order fringe will be $1.25mm + 1.4mm = 1.65mm$
And as $n = 0$ in ${y_n} = \dfrac{{n\lambda D}}{d}$ obviously the position of the zero order fringe remains unchanged and equal to $1.25mm$ .

$\therefore$ Adding the zero error of the micrometer gives the new position of ${10^{th}}$ order fringe will be $1.65mm$ and the position of the zero-order of the fringe remains unchanged. Hence the correct option is (A).

Note:
Always remember that when the position of zero-order fringe is not zero, in this type of experiment, then there must be a zero error in the micrometer. Fringe width is independent of the order of fringe. Fringe width is directly proportional to the wavelength of the light used.