Question

A screen is at a distance D = 80 cm from a diaphragm having two narrow slits $S_1$ and $S_2$ which are d = 2 mm apart. Slit $S_1$ is covered by a transparent sheet of thickness $t_1$ = 2.5 $\mu$m and $S_2$ by another sheet of thickness $t_2$ = 1.25 $\mu$m as shown in figure. Both sheets are made of same material having refractive index $\mu$ = 1.40. Water is filled in space between diaphragm and screen. A monochromatic light beam of wavelength $\lambda$ = 5000 $\mathring{A}$ is incident normally on the diaphragm. Assuming intensity of beam to be uniform and slits of equal width, calculate ratio of intensity at C to maximum intensity of interference pattern obtained on the screen, where C is foot of perpendicular bisector of $S_1 S_2$. (Refractive index of water, $\mu_w$ = 4/3)

Answer 1

$\cos^2(40^\circ)$

Answer 2

The given parameters are:

• Distance between screen and diaphragm, $D = 80 \text{ cm} = 0.8 \text{ m}$.

• Distance between the two slits, $d = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$.

• Thickness of the transparent sheet covering slit $S_1$, $t_1 = 2.5 \text{ } \mu\text{m} = 2.5 \times 10^{-6} \text{ m}$.

• Thickness of the transparent sheet covering slit $S_2$, $t_2 = 1.25 \text{ } \mu\text{m} = 1.25 \times 10^{-6} \text{ m}$.

• Refractive index of the material of the sheets, $\mu = 1.40$.

• Refractive index of water, $\mu_w = 4/3$.

• Wavelength of the incident light in vacuum, $\lambda = 5000 \text{ } \mathring{A} = 5000 \times 10^{-10} \text{ m} = 5 \times 10^{-7} \text{ m}$.

The wavelength of light in water is $\lambda_w = \frac{\lambda}{\mu_w} = \frac{5 \times 10^{-7} \text{ m}}{4/3} = \frac{15}{4} \times 10^{-7} \text{ m} = 3.75 \times 10^{-7} \text{ m}$.

When a transparent sheet of thickness $t$ and refractive index $\mu$ is placed in a medium of refractive index $\mu_w$, the additional optical path introduced compared to the path in the medium is $(\mu - \mu_w)t$.

For slit $S_1$, the additional optical path is $\Delta L_1 = (\mu - \mu_w)t_1$.

For slit $S_2$, the additional optical path is $\Delta L_2 = (\mu - \mu_w)t_2$.

At point C, which is the foot of the perpendicular bisector of $S_1S_2$, the geometrical path difference is zero. The optical path difference at C is due to the presence of the transparent sheets:

$\Delta L = \Delta L_1 - \Delta L_2 = (\mu - \mu_w)t_1 - (\mu - \mu_w)t_2 = (\mu - \mu_w)(t_1 - t_2)$.

Calculate $\mu - \mu_w$: $\mu - \mu_w = 1.40 - 4/3 = \frac{14}{10} - \frac{4}{3} = \frac{7}{5} - \frac{4}{3} = \frac{21 - 20}{15} = \frac{1}{15}$.

Calculate $t_1 - t_2$: $t_1 - t_2 = 2.5 \times 10^{-6} \text{ m} - 1.25 \times 10^{-6} \text{ m} = 1.25 \times 10^{-6} \text{ m} = \frac{5}{4} \times 10^{-6} \text{ m}$.

Calculate the optical path difference $\Delta L$: $\Delta L = \frac{1}{15} \times \frac{5}{4} \times 10^{-6} \text{ m} = \frac{1}{3 \times 4} \times 10^{-6} \text{ m} = \frac{1}{12} \times 10^{-6} \text{ m}$.

The phase difference $\phi$ at point C is given by $\phi = \frac{2\pi}{\lambda_w} \Delta L$. $\phi = \frac{2\pi}{(15/4) \times 10^{-7} \text{ m}} \times \frac{1}{12} \times 10^{-6} \text{ m} = \frac{8\pi}{15 \times 10^{-7}} \times \frac{1}{12} \times 10^{-6} = \frac{8\pi}{15 \times 12} \times 10 = \frac{80\pi}{180} = \frac{4\pi}{9}$.

Assuming the intensity of light from each slit is $I_0$ (since the intensity of the beam is uniform and slits are of equal width), the intensity at point C is given by $I_C = 4I_0 \cos^2 (\phi/2)$.

The maximum intensity in the interference pattern is $I_{max} = 4I_0$.

The ratio of intensity at C to the maximum intensity is $\frac{I_C}{I_{max}} = \frac{4I_0 \cos^2 (\phi/2)}{4I_0} = \cos^2 (\phi/2)$.

Substitute the value of $\phi$: $\phi/2 = \frac{1}{2} \times \frac{4\pi}{9} = \frac{2\pi}{9}$.

The ratio is $\cos^2 \left(\frac{2\pi}{9}\right)$.

To express this numerically, $\frac{2\pi}{9}$ radians is equal to $\frac{2 \times 180^\circ}{9} = 40^\circ$.

The ratio is $\cos^2 (40^\circ)$.