In a Young’s double-slit experiment, two slits are 1.5 mm apart while the screen is 1.2 m away. When a light of wavelength 600 nm is incident on slits, the fringe width will be

The correct option is 0.48 mm.

To find the fringe width in Young's double-slit experiment, we use the standard formula for fringe width ($\beta$):

$$\beta =\frac{\lambda D}{d}$$

Where:
• $\lambda$ is the wavelength of the light source.
• $D$ is the distance between the slits and the screen.
• $d$ is the distance between the two slits.

From the given problem, we have the following values:
• Slit separation, $d=1.5\text{ mm}=1.5\times {10}^{-3}\text{ m}$
• Distance to the screen, $D=1.2\text{ m}$
�� Wavelength of light, $\lambda =600\text{ nm}=600\times {10}^{-9}\text{ m}=6\times {10}^{-7}\text{ m}$

Now, let's substitute these values into the fringe width formula:

$$\beta =\frac{(6\times {10}^{-7}\text{ m})\times \left(1.2\text{ m}\right)}{1.5\times {10}^{-3}\text{ m}}$$

Simplify the numerator:
$$6\times 1.2=7.2$$

Substitute this back into the equation:

$$\beta =\frac{7.2\times {10}^{-7}}{1.5\times {10}^{-3}}$$

Now, perform the division:
$$\frac{7.2}{1.5}=4.8$$

And combine the powers of 10:
$${10}^{-7-(-3)}={10}^{-4}$$

Thus, we get:

$$\beta =4.8\times {10}^{-4}\text{ m}$$

To convert the unit from meters (m) to millimeters (mm), multiply by 1000 (${10}^3$):

$$\beta =4.8\times {10}^{-4}\times {10}^3\text{ mm}$$

$$\beta =4.8\times {10}^{-1}\text{ mm}=0.48\text{ mm}$$

Therefore, the fringe width is 0.48 mm.