Use the formula for bright fringes in YDSE

The position of the $n^{\text{th}}$ bright fringe from the central maximum is $y_n = n\beta = n\frac{\lambda D}{d}$ where:

• $y_n$ = position of $n^{\text{th}}$ bright fringe
• $\lambda$ = wavelength
• $D$ = distance between slits and screen
• $d$ = slit separation

Given data

• $y_5 = 5\text{ cm} = 5 \times 10^{-2}\text{ m}$
• $D = 1\text{ m}$
• $\lambda = 600\text{ nm} = 600 \times 10^{-9}\text{ m}$
• $n = 5$

Substitute into the formula

$y_5 = 5\frac{\lambda D}{d}$

So, $5 \times 10^{-2} = 5\frac{600 \times 10^{-9} \times 1}{d}$

Solve for $d$

$d = 5\frac{600 \times 10^{-9}}{5 \times 10^{-2}}$

Cancel $5$: $d = \frac{600 \times 10^{-9}}{10^{-2}}$

$d = 600 \times 10^{-7} \text{ m}$

$d = 6 \times 10^{-5} \text{ m}$

Convert into micrometers

$6 \times 10^{-5} \text{ m} = 60 \times 10^{-6} \text{ m} = 60\,\mu\text{m}$

Check options

The correct option is: $\boxed{A:~60\,\mu\text{m}}$