Answer to Question #93477 in Physics for Shivam Manishrao Hiware

Question

In youngs double slit experiment, the slits are 0.5 mm apart and interference is observed on a screen placed at a distance of 100 cm from the slits. It is found that 9 th bright fringe is at a distance 8.835 mm from the second dark fringe on the same side of the centre of the fringe pattern. Find the wavelength of light used?

Answer 1

Let us use notation: $d = 0.5 mm$ - distance between the slits, $D = 1m$ - distance from the slits to the screen, $l_9$ - distance from the center of the pattern to 9^th maximum, $l_2$ - distance from the center of the pattern to 2^nd minimum.

The condition for the m^th maximum is $d \sin \theta = m \lambda$ , and for n^th minimum is $d \sin \theta = (n + \frac{1}{2}) \lambda$, where $\lambda$ is the wavelength. Since the distance from the slit to the screen is quite big, let us use approximation $\sin \theta \approx \tan \theta = \frac{y}{D}$ , where $y$ is the distance from the center of the pattern to particular maximum/minimum. Hence, one has equations $\frac{d l_9}{D} = 9 \lambda$ , $\frac{d l_2}{D} = (2+ \frac{1}{2}) \lambda$ for maximum and minimum respectively. Subtracting the second equation from the first, obtain $\frac{d}{D}(l_9 - l_2) = 7.5 \lambda$. The distance between the maximum and minimum is given, $l_9 - l_2 = 8.835 mm$, therefore the wavelength is $\lambda = \frac{2 d (l_9 - l_2)} {15 D} = 589 nm$.