Single-slit diffraction: A laser beam passes through a thin slit. When the pattern is viewed on a screen 1.25 m past the slit, you observe that the fifth-order dark fringes occur at ±2.41 cm from the central bright fringe. The entire experiment is now performed within a liquid, and you observe that each of the fifth-order dark fringes is 0.790 cm closer to the central fringe than it was in air. What is the index of refraction of this liquid?
A) 1.33
B) 1.40
C) 1.49
D) 1.62
E) 3.05

Question

Quizplus · Accepted Answer

The position of the dark fringes in air is given by:
$$x_n = n\frac{\lambda L}{d}$$
where $n$ is the order of the fringe, $\lambda$ is the wavelength of the laser, $L$ is the distance from the slit to the screen, and $d$ is the width of the slit.

For the fifth-order fringe, we have:
$$2x_5 = 2\cdot5\frac{\lambda L}{d} = 0.0241	ext{ m}$$
In the liquid, the wavelength of the laser is reduced by a factor of the refractive index $n$ of the liquid:
$$\lambda' = \frac{\lambda}{n}$$
The distance between fringes is still given by $x_n = n\frac{\lambda' L}{d}$. For the fifth-order fringe, we have:
$$2x_5' = 2\cdot5\frac{\lambda' L}{d} = 0.0169	ext{ m}$$
The difference in position between the fifth-order fringes in air and in the liquid is:
\begin{align*}
\Delta x_5 &= x_5'-x_5 \
&= 0.0241	ext{ m} - 0.0169	ext{ m} \
&= 0.0072	ext{ m}
\end{align*}
This is equal to the amount by which each fringe has moved closer to the central fringe:
$$\Delta x_5 = 0.790	ext{ cm} = 0.0079	ext{ m}$$
Therefore, we have:
$$\frac{\Delta x_5}{x_5} = \frac{0.0072	ext{ m}}{0.0241	ext{ m}} = \frac{0.0079	ext{ m}}{0.0241	ext{ m}-0.0079	ext{ m}} = n-1$$
Solving for $n$, we get:
$$n \approx 1.49$$
Therefore, the index of refraction of the liquid is approximately 1.49. The best choice is (C).

Single-Slit Diffraction: a Laser Beam Passes Through a Thin Slit