In a double-slit experiment, monochromatic light of wavelength 500\text{ nm} is incident n… — JAMB UTME

In a double-slit experiment, monochromatic light of wavelength $500\text{ nm}$ is incident normally on two narrow slits. If the third-order ( $m = 3$ ) bright fringe is observed at an angle of $30^\circ$ from the central maximum, what is the slit separation, $d$ , in micrometers ( $\mu\text{m}$ )?

Answer: 3 µm

Answer

The slit separation is

3.0\text{ }\mu\text{m}

For bright fringes in a double-slit setup, constructive interference occurs when

d \sin\theta = m\lambda

. Rearranging for slit separation yields

d = \frac{m\lambda}{\sin\theta}

. Substituting

m = 3

\lambda = 0.50\text{ }\mu\text{m}

, and

\theta = 30^\circ

gives

d = \frac{3 \times 0.50\text{ }\mu\text{m}}{\sin 30^\circ} = \frac{1.50\text{ }\mu\text{m}}{0.50} = 3.0\text{ }\mu\text{m}

Step-by-Step Solution

Identify the constructive interference equation for Young's double-slit experiment.

d \sin\theta = m\lambda

Bright fringes occur where waves from the two slits interfere constructively, which corresponds to path differences equal to integer multiples of the wavelength.

Convert the given wavelength to micrometers and substitute known values.

\lambda = 0.50\text{ }\mu\text{m}

m = 3

\sin(30^\circ) = 0.50

Converting units to micrometers early simplifies direct calculation of

d

\mu\text{m}

Rearrange the equation and evaluate for

d

d = \frac{3 \times 0.50\text{ }\mu\text{m}}{0.50} = 3.0\text{ }\mu\text{m}

Dividing the numerator by

\sin(30^\circ) = 0.50

doubles the value of

m\lambda

Key Concept

Angular position condition for constructive interference in double-slit diffraction

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