Question 309: Waves Optics

Question

In an interference experiment, the $\mathrm{n}^{	ext {th }}$ bright fringe for light of wavelength $\lambda_1(\mathrm{n}=0,1,2,3 \ldots)$ coincides with the $\mathrm{m}^{	ext {th }}$ dark fringe for light of wavelength $\lambda_2(\mathrm{~m}=1,2,3 \ldots)$. The ratio $\frac{\lambda_1}{\lambda_2}$ is

Accepted Answer

In an interference experiment, the condition for a bright fringe due to a light of wavelength $\lambda_1$ is given by the equation:
$$ d \sin 	heta = n \lambda_1, $$
where $d$ is the slit separation, $	heta$ is the angle of diffraction, and $n$ is the order of the bright fringe ($n = 0, 1, 2, \ldots$).
For a dark fringe due to a light of wavelength $\lambda_2$, the condition is:
$$ d \sin 	heta = \left(m - \frac{1}{2}ight) \lambda_2, $$
where $m$ is the order of the dark fringe ($m = 1, 2, 3, \ldots$).
If both these fringes coincide, their path differences must be equal. Therefore:
$$ n \lambda_1 = \left(m - \frac{1}{2}ight) \lambda_2. $$
Solving for the ratio $\frac{\lambda_1}{\lambda_2}$, we have:
$$ \frac{\lambda_1}{\lambda_2} = \frac{m - \frac{1}{2}}{n}. $$
Multiplying the numerator and the denominator by 2 to clear the fraction, we obtain:
$$ \frac{\lambda_1}{\lambda_2} = \frac{2m - 1}{2n}. $$
So, the correct option is
Option C
: $\frac{2m - 1}{2n}$.

Answer

$\frac{\mathrm{m}-1}{\mathrm{n}}$

Answer

$\frac{2 \mathrm{~m}-1}{\mathrm{n}}$

Answer

$\frac{2 m-1}{2 n}$

Answer

$\frac{2 \mathrm{~m}+1}{2 \mathrm{n}}$

Waves Optics

In an interference experiment, the \mathrm{n}^{\text {th }} bright fringe for light of wavelength \lambda_1(\mathrm{n}=0,1,2,3 \ldots) coincides with the \mathrm{m}^{\text {th }} dark fringe for light of wavelength \lambda_2(\mathrm{~m}=1,2,3 \ldots). The ratio \frac{\lambda_1}{\lambda_2} is

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