Question 32: Wave Optics

Question

In Young's double slit experiment the ratio of phase difference between light waves reaching the third bright fringe and third dark fringe is

Accepted Answer

In Young's double slit experiment:
Ã¢ÂÂ¢ Bright fringes occur when the path difference is an integer multiple of the wavelength. Mathematically, the condition is:
$$ d\sin	heta = m\lambda \quad (m = 0, 1, 2, \ldots) $$
Ã¢ÂÂ¢ Dark fringes occur when the path difference is an odd multiple of half the wavelength, i.e.,
$$ d\sin	heta = \left(m+\frac{1}{2}ight)\lambda \quad (m = 0, 1, 2, \ldots) $$
Now, letÃ¢ÂÂs determine the phase differences for the fringes in question.
For any fringe, the phase difference is related to the path difference by:
$$    \phi = \frac{2\pi}{\lambda} 	imes (	ext{path difference})    $$
For the third bright fringe (taking $m=3$):
The path difference is:
$$      \Delta_{	ext{bright}} = 3\lambda      $$
Therefore, the phase difference is:
$$      \phi_{	ext{bright}} = \frac{2\pi}{\lambda} 	imes 3\lambda = 6\pi      $$
For the third dark fringe (with $m=2$; since the dark fringes occur at $m+\frac{1}{2}$, the third dark fringe corresponds to $m=2$):
The path difference is:
$$      \Delta_{	ext{dark}} = \left(2+\frac{1}{2}ight)\lambda = \frac{5\lambda}{2}      $$
Thus, the phase difference is:
$$      \phi_{	ext{dark}} = \frac{2\pi}{\lambda} 	imes \frac{5\lambda}{2} = 5\pi      $$
Now, the ratio of the phase differences between the third bright fringe and the third dark fringe is:
$$    \frac{\phi_{	ext{bright}}}{\phi_{	ext{dark}}} = \frac{6\pi}{5\pi} = \frac{6}{5}    $$
This corresponds to Option C: $$\frac{6}{5}.$$

Answer

$$\frac{4}{3}$$

Answer

$$\frac{5}{2}$$

Answer

$$\frac{6}{5}$$

Answer

$$\frac{7}{6}$$

Wave Optics

In Young's double slit experiment the ratio of phase difference between light waves reaching the third bright fringe and third dark fringe is

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