# chapt36 - Chapter 36 Diffraction Problem 3 Light of...

This preview shows pages 1–4. Sign up to view the full content.

Chapter 36: Diffraction Problem 3 Light of wavelength 633 nm is incident on a narrow slit. The angle between the first diffraction minimum on one side of the central maximum and the first minimum on the other side is 1.20 ° . What is the width of the slit? Answer 60.4 μ m. Solution The fact that there is an angle of 1.20 ° between the two minima means that the angle to either minimum from the center is 0.60 ° . The single-slit diffraction minima are located according to the equation a sin( θ ) = m λ , for m = 1, 2, 3, . .. Therefore the slit width in this case is a = m λ sin θ ( ) = 1 633 nm sin 0.60 ° ( ) = 60.4 μ m Problem 5 A plane wave of wavelength 590 nm is incident on a slit with a width of a = 0.40 mm. A thin converging lens of focal length +70 cm is placed between the slit and a viewing screen and focuses the light on the screen. (a) How far is the screen from the lens? (b) What is the distance on the screen from the center of the diffraction pattern to the first minimum? Answer (a) 70 cm; (b) 1.0 mm Solution (a) The rays passing through the slit are essentially parallel after passing through the slit therefore they will be focused by the lens at the focal length of the lens, 70 cm. (b) The angle, θ , to the first minimum is given by sin( θ ) = λ /a If the distance to the first minimum is y and the distance to the screen is the focal length, f, then (see Diagram 36-5) tan( θ ) = y/f

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
36-2 Chapter 36: Diffraction θ f y Diagram 36-8 For small angles sin( θ ) = tan( θ ). Therefore y = f λ a = 70 cm 590 nm 0.40 mm = 1.0 mm Problem 7 The distance between the first and fifth minima of a single-slit diffraction pattern is 0.35 mm with the screen 40 cm away from the slit, when light of wavelength 550 nm is used. (a) Find the slit width. (b) Calculate the angle θ of the first diffraction minimum. Answer (a) 2.5 mm; (b) 0.000219 rad Solution (a) The location of the m th minimum is given by a sin( θ ) = m λ If the screen is a distance D from the slit and the distance from the central maximum to the m th minimum is y m then (refer to Diagram 36-8 for a similar situation) tan( θ ) = y m /D For the typically small angles involved tan( θ ) sin( θ ) therefore y m = m D a and the distance between the first and fifth minima is Δ y = y 5 y 1 = 5 D a D a = 4 D a
Chapter 36: Diffraction 36-3 Therefore the slit width is a = 4 λ D Δ y = 4 550 nm 40 cm 0.35 mm = 2.5 mm Problem 11 Monochromatic light with wavelength 538 nm is incident on a slit with width 0.025 mm. The distance from the slit to a screen is 3.5 m. Consider a point on the screen 1.1 cm from the central maximum. Calculate (a) θ for that point, (b) α , and (c) the ratio of the intensity at this point to the intensity at the central maximum.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 08/03/2009 for the course PHYSICS 2049 taught by Professor C during the Spring '09 term at Vanderbilt.

### Page1 / 14

chapt36 - Chapter 36 Diffraction Problem 3 Light of...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online