For an experiment using Foucault's apparatus, the setup distances...

In Foucault's rotating mirror experiment, the shift in the image is given by the formula:

$s = \dfrac{4 R^2 \omega a}{c (R+b)}$

where:
$R$ = distance between the rotating and fixed mirror = $16 \text{ m}$
$b$ = distance between the lens and the rotating mirror = $6 \text{ m}$
$a$ = distance between the source and the lens = $2 \text{ m}$
$\omega$ = angular velocity of the rotating mirror = $2 \pi n$
$n$ = frequency of rotation = $356 \text{ rev/s}$
$s$ = shift in the image = $0.7 \text{ mm} = 0.7 \times 10^{-3} \text{ m}$
$c$ = speed of light

Rearranging the formula to solve for the speed of light $c$:

$c = \dfrac{4 R^2 (2 \pi n) a}{s (R+b)}$

Substituting the given values and using $\pi \approx \dfrac{22}{7}$:

$c = \dfrac{8 \times \dfrac{22}{7} \times 356 \times (16)^2 \times 2}{0.7 \times 10^{-3} \times (16 + 6)}$

$c = \dfrac{8 \times 22 \times 356 \times 256 \times 2}{7 \times 0.7 \times 10^{-3} \times 22}$

Canceling $22$ from the numerator and denominator:

$c = \dfrac{8 \times 356 \times 256 \times 2}{4.9 \times 10^{-3}}$

$c = \dfrac{4096 \times 356}{4.9} \times 10^3$

$c = \dfrac{1458176}{4.9} \times 10^3$

$c \approx 297586 \times 10^3 \text{ m/s}$

$c \approx 2.9758 \times 10^8 \text{ m/s}$

Rounding to two decimal places, we get $c \approx 2.98 \times 10^8 \text{ m/s}$.

Answer: $2.98 \times 10^8 \text{ m/s}$