Question

A mica strip and a polysterene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of mica and polysterene are 1.58 and 1.55 respectively for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen a distance one meter away.

(a) What would be the fringe-width?

(b) At what distance from the centre will the first maximum be located?

Answer 1

(a) 0.492 mm (b) 12.99 mm

Answer 2

(a) The fringe-width ($\beta$) in a double-slit interference setup is given by the formula $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength of light, $D$ is the distance from the slits to the screen, and $d$ is the separation between the slits. The presence of the thin strips shifts the fringe pattern but does not change the fringe-width, as the wavelength of light in the region where interference occurs (air) remains $\lambda$. Substitute the given values $\lambda = 590 \times 10^{-9}$ m, $D = 1$ m, and $d = 0.12 \times 10^{-2}$ m into the formula to calculate $\beta$.

(b) The thin strips placed in front of the slits introduce an additional optical path difference. If a strip of thickness $t$ and refractive index $\mu$ is placed in front of a slit, the optical path through the strip is $\mu t$, which is equivalent to a geometric path of $\mu t$ in vacuum. Compared to the path through air of the same thickness $t$ (optical path $t$), the strip introduces an extra optical path of $(\mu - 1)t$. If strips of refractive indices $\mu_1$ and $\mu_2$ and thickness $t$ are placed in front of the two slits, the total optical path difference between the waves from the two slits reaching a point P on the screen at a distance $y$ from the center is $\Delta = \frac{yd}{D} + (\mu_2 - \mu_1)t$ (assuming appropriate assignment of indices and $y$ direction). For a bright fringe (maximum), $\Delta = n\lambda$, where $n$ is an integer. Thus, $\frac{yd}{D} + (\mu_2 - \mu_1)t = n\lambda$. The position of the $n$-th maximum is $y_n = \frac{D}{d} [n\lambda + (\mu_1 - \mu_2)t]$. The central maximum is at $n=0$, so its position is $y_0 = \frac{D}{d}(\mu_1 - \mu_2)t$. The first maximum corresponds to $n=1$. Its position is $y_1 = \frac{D}{d} [\lambda + (\mu_1 - \mu_2)t] = y_0 + \beta$. Assuming mica ($\mu_1 = 1.58$) is in front of one slit and polysterene ($\mu_2 = 1.55$) is in front of the other, calculate $y_0$ using the given values of $D, d, \mu_1, \mu_2, t$. Then calculate $y_1 = y_0 + \beta$. The distance from the center is $|y_1|$.

Answer: (a) The fringe-width is calculated as: $\beta = \frac{(590 \times 10^{-9} \text{ m}) \times (1 \text{ m})}{0.12 \times 10^{-2} \text{ m}} = \frac{590 \times 10^{-7}}{0.12} \text{ m} = \frac{59000}{12} \times 10^{-7} \text{ m} = \frac{14750}{3} \times 10^{-7} \text{ m} = 4916.67 \times 10^{-7} \text{ m} = 0.491667 \times 10^{-3} \text{ m} = 0.491667$ mm. Rounding to three significant figures, the fringe-width is 0.492 mm.

(b) The position of the central maximum ($n=0$) is $y_0 = \frac{D}{d}(\mu_1 - \mu_2)t$. Let $\mu_1 = 1.58$ (mica) and $\mu_2 = 1.55$ (polysterene). $y_0 = \frac{1 \text{ m}}{0.12 \times 10^{-2} \text{ m}} \times (1.58 - 1.55) \times (0.50 \times 10^{-3} \text{ m})$ $y_0 = \frac{1}{0.0012} \times 0.03 \times 0.0005 \text{ m} = \frac{0.000015}{0.0012} \text{ m} = \frac{15 \times 10^{-6}}{12 \times 10^{-4}} \text{ m} = 1.25 \times 10^{-2} \text{ m} = 0.0125 \text{ m} = 12.5$ mm. The position of the first maximum ($n=1$) from the center is $y_1 = y_0 + \beta$. $y_1 = 12.5 \text{ mm} + 0.491667 \text{ mm} = 12.991667 \text{ mm}$. Rounding to three significant figures, the distance of the first maximum from the center is 12.99 mm.