Question 14

The position of the $n^{th}$ bright fringe on the screen is given by $y_n = \frac{n \lambda D}{d}$.

Given parameters:

• $n = 8$
• $\lambda = 6000$ \AA $= 6 \times 10^{-7}$ m
• $D = 1$ m
• $d(t) = (0.8 + 0.04 \sin \omega t) \times 10^{-3}$ m

The separation between the slits $d$ oscillates between:

• $d_{min} = 0.8 - 0.04 = 0.76$ mm $= 0.76 \times 10^{-3}$ m
• $d_{max} = 0.8 + 0.04 = 0.84$ mm $= 0.84 \times 10^{-3}$ m

The extreme positions of the $8^{th}$ bright fringe are: $y_{max} = \frac{8 \lambda D}{d_{min}} = \frac{8 \times 6 \times 10^{-7} \times 1}{0.76 \times 10^{-3}} \approx 6.3157 \times 10^{-3}$ m $y_{min} = \frac{8 \lambda D}{d_{max}} = \frac{8 \times 6 \times 10^{-7} \times 1}{0.84 \times 10^{-3}} \approx 5.7143 \times 10^{-3}$ m

The separation between these extreme positions is: $\Delta y = y_{max} - y_{min} = 8 \lambda D \left( \frac{1}{d_{min}} - \frac{1}{d_{max}} \right)$ $\Delta y = 48 \times 10^{-4} \left( \frac{0.84 - 0.76}{0.76 \times 0.84} \right) = 48 \times 10^{-4} \left( \frac{0.08}{0.6384} \right)$ $\Delta y \approx 6.015 \times 10^{-4} \text{ m} = 601.5 \text{ } \mu\text{m}$

Rounding to two decimal places as per general instructions, we get $601.50$.