H2 Physics 11 - Superposition: Two-Slit Interference Pattern

Step 1 - State the problem and recall the condition for interference

A double-slit experiment uses light of wavelength five hundred and forty nanometres.

Step 1 - State the problem and recall the condition for interference

The slit separation is zero point four millimetres and the screen is one point five metres away.

Step 1 - State the problem and recall the condition for interference

We want to find the fringe spacing on the screen and identify the position of the third bright fringe.

Step 2 - Apply the fringe spacing formula

For double-slit interference the fringe spacing x equals lambda D over d.

Step 2 - Apply the fringe spacing formula

This formula assumes the slit separation d is much smaller than the distance D to the screen, which holds here.

Step 2 - Apply the fringe spacing formula

Substituting: x equals five point four zero times ten to the minus seven, times one point five, divided by four point zero times ten to the minus four.

Step 3 - Evaluate the fringe spacing

Multiplying the numerator gives eight point one zero times ten to the minus seven.

Step 3 - Evaluate the fringe spacing

Dividing by four point zero times ten to the minus four gives two point zero three times ten to the minus three metres.

Step 3 - Evaluate the fringe spacing

That is about two point zero millimetres per fringe.

Step 4 - Find the position of the third bright fringe

Bright fringes occur where the path difference equals n lambda, with n equal to zero, one, two, three and so on.

Step 4 - Find the position of the third bright fringe

The position of the n-th bright fringe from the central maximum is y equals n times x.

Step 4 - Find the position of the third bright fringe

For the third bright fringe, y equals three times two point zero three millimetres, which gives six point one millimetres.

Step 5 - Discuss what happens if the slit separation is halved

If the slit separation d is halved to zero point two millimetres, the fringe spacing doubles to about four millimetres.

Step 5 - Discuss what happens if the slit separation is halved

This is because fringe spacing is inversely proportional to d.

Step 5 - Discuss what happens if the slit separation is halved

A common mistake is to confuse slit separation d with slit width a.

Step 5 - Discuss what happens if the slit separation is halved

Remember: d controls the fringe spacing, while a controls the diffraction envelope that modulates the intensity pattern.