Light with a wavelength of 600.0 nm is incident on a singleslit. The resulting d

Question

Light with a wavelength of 600.0 nm is incident on a singleslit. The resulting diffraction pattern is observed on a screenthat is 0.50 m from the slit. The distance between the first andthird minima of the diffraction pattern is 0.80 mm. Which range ofvalues listed below contains the width of the slit?

a. 0.4 mm to 0.8 mm
b. 1.2 mm to 1.6 mm
c. 0.1 mm to 0.4 mm
d. 0.8 mm to 1.2 mm
e. 1.6 mm to 2.0 mm Light with a wavelength of 600.0 nm is incident on a singleslit. The resulting diffraction pattern is observed on a screenthat is 0.50 m from the slit. The distance between the first andthird minima of the diffraction pattern is 0.80 mm. Which range ofvalues listed below contains the width of the slit? 0.4 mm to 0.8 mm 1.2 mm to 1.6 mm 0.1 mm to 0.4 mm 0.8 mm to 1.2 mm 1.6 mm to 2.0 mm

Explanation / Answer

the general condition for destructive interference is sindark = m * (/a) m =±1,±2,±3,.............. -----------(1) This equation gives the values of dark forwhich the diffraction pattern has zero intensity,that is,when adark fringe is formed. when m = 1 then sin1 = 1 * (/a1) or a1 = (/sin1)-----------(1) when m = 3 then sin3 = 1 * (/a3) or a3 = (/sin3)------------(2) the first minima measured from the central axis is y1 = L * sindark = L *sin1 similarly,the third minima is y3 = L * sin3 the distance between the first and third minima of thediffraction pattern is 0.80 mm that is y3 - y1 = L * sin3 - L* sin1 = L * (sin3 -sin1) or L * (sin3 - sin1) = 0.80* 10-3 or (sin3 - sin1) = (0.80 *10-3/L) L = 0.50 m or (sin3 - sin1) = (0.80 *10-3/0.50) or  (sin3 - sin1)= 1.6 * 10-3 or sin3 = sin1 + 1.6 *10-3 ------------(3) from equations (2) and (3) we get a3 = (/sin1 + 1.6 *10-3) therefore,the range of values which contains the width of theslit is a3 = (/sin1 + 1.6 *10-3) and a1 =(/sin1) = 600.0 nm = 600.0 * 10-9 m the range of values which contains the width of the slit isinversely proportional to the sine of the angle1. or sin3 = sin1 + 1.6 *10-3 ------------(3) from equations (2) and (3) we get a3 = (/sin1 + 1.6 *10-3) therefore,the range of values which contains the width of theslit is a3 = (/sin1 + 1.6 *10-3) and a1 =(/sin1) = 600.0 nm = 600.0 * 10-9 m the range of values which contains the width of the slit isinversely proportional to the sine of the angle1.

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

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