Monochromatic light of wavelength 355 nm is incident on a narrow slit. On a scre

Question

Monochromatic light of wavelength 355 nm is incident on a narrow slit. On a screen 3.00 m away, the distance between the second diffraction minimum and the central maximum is 1.50 cm. Calculate the angle of diffraction theta of the second minimum. Find the width of the slit. The format we use for Diffraction by a single slit a sin (pheta) = m lambda Where a is the slit width, pheta the angle between the central max and the desired minima, m is the number of the minima, and lambda the wavelength. I was able to find the angle just fine but when plugging that back into the formula to solve for a... I can't seem to get the right answer

Explanation / Answer

a sin theta = (m+1/2) * wavelength

a = 3/2* 355*10^-9 / sin(0.2865)   (here, second minimum, so, m=1)

a = 106.492*10^-6 m = 0.106 mm

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

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