In a double-slit lab experiment, two slits are known to be 2.00 mm apart. A mixt

Question

In a double-slit lab experiment, two slits are known to be 2.00 mm apart. A mixture of two wavelengths of light shines simultaneously on the slits and this light is diffracted though the slits. The wavelengths are 1 = 750.0 nm and 2 = 900.0 nm. The two interference patterns share a common central maximum. On a screen positioned 2.00 m from the slits, at what minimum distance from the central maximum will a bright fringe of one wavelength light coincide (overlap exactly) with a bright fringe of the other wavelength light?

Explanation / Answer

The formula that applies here is y/L = m(wavelength)/d

For the first wavelength, we will calculate with a few m values.

For m = 1

y/2 = (1)(750 X 10-9)/(2 X 10-3)

y = 7.5 X 10-4 m

For m = 2...y = 1.5 X 10-3 m

For m = 3... y = 2.25 X 10-3 m

For m = 4... y = 3 X 10-3 m

For m = 5... y = 3.75 X 10-3 m

For m = 6... y = 4.5 X 10-3 m

Do the same for the 900 nm wavelength...

When m = 1

y/2 = (1)(900 X 10-9)/(2 X 10-3)

y = 9 X 10-4 m

When m = 2... y = 1.8 X 10-3 m

When m = 3... y = 2.7 X 10-3 m

When m = 4... y = 3.6 X 10-3 m

When m = 5... y = 4.5 X 10-3 m

Thus for 750 nm, m will be 6 and for 900 nm, m will be 5.

The two will overlap at 4.5 X 10-3 m (4.5 mm)

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.