Two slits separated by a distance of d = 0.155 mm are located at a distance of D

Question

Two slits separated by a distance of d = 0.155 mm are located at a distance of D = 2.14 m from a screen. The screen is oriented parallel to the plane of the slits. The slits are illuminated by a monochromatic and coherent light source with a wavelength of lambda = 608 nm. A wave from each slit propagates to the screen. The interference pattern shows a peak at the center of the screen (m = 0) and then alternating minima and maxima. What is the path length difference between the waves at the first maximum (m = 1) on the screen? At what angle from the beam axis will the first (m = 1) maximum appear? (You can safely use the small angle approximation.)

Explanation / Answer

d = 0.155 mm ; D = 2.14 m ; lambda = 608 nm

a)for m = 1,

the path difference would be = 1 x lambda

pd = 608 nm

b)We know that

d sin(theta) = m lambda

theta = sin -1 [m lambda/d]

theta = sin -1 [1 x 608 x 10^-9/0.155 x 10^-3] = 0.225 deg

Hence, theta = 0.225 deg

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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