In a Young's double-slit experiment, the separation between two narrow slits is 0.40\text{… — JAMB UTME

In a Young's double-slit experiment, the separation between two narrow slits is $0.40\text{ mm}$ and the interference pattern is observed on a screen placed $1.20\text{ m}$ away from the slits. If the distance between consecutive bright fringes on the screen is $1.80\text{ mm}$ , what is the wavelength of the light used in nanometers ( $\text{nm}$ )?

Answer: 600 nm

Answer

The wavelength of the light used is

600\text{ nm}

Using the Young's double-slit fringe spacing relation

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

, rearranging yields

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

. Substituting

\beta = 1.80 \times 10^{-3}\text{ m}

d = 4.0 \times 10^{-4}\text{ m}

, and

D = 1.20\text{ m}

gives

\lambda = 6.00 \times 10^{-7}\text{ m}

, which corresponds to

600\text{ nm}

Step-by-Step Solution

Convert given physical quantities into standard SI units (meters).

Slit separation

d = 0.40\text{ mm} = 4.0 \times 10^{-4}\text{ m}

, distance to screen

D = 1.20\text{ m}

, and fringe spacing

\beta = 1.80\text{ mm} = 1.80 \times 10^{-3}\text{ m}

Standard SI units ensure accuracy when applying wave speed and distance equations.

Write the Young's double-slit formula relating fringe width to wavelength.

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

This relationship defines the spatial period of interference fringes on a screen.

Rearrange the equation to isolate the wavelength

\lambda

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

The unknown parameter to solve for is the wavelength of the monochromatic source.

Substitute the numerical values and convert the final result to nanometers.

$\lambda = \frac{1.80 \times 10^{-3}\text{ m} \times 4.0 \times 10^{-4}\text{ m}}{1.20\text{ m}} = 6.00 \times 10^{-7}\text{ m} = 600\text{ nm}$

Multiply meters by

10^9

to express the wavelength in nanometers.

Key Concept

Young's Double-Slit Interference Fringe Spacing

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