Ch 25 Homework

Ch 25: 5, 17, 28, 40, 44, 62, 63

25.5 Two slits 0.03 mm apart are illuminated by parallel rays and the 5th-order maximum is 14 cm from the central bright maximum on a screen 2.0 m from the slits. What is the wavelength of the light falling on the slits?

Maxima (or bright fringes) occur for

d sin = m

We can solve this for

= (d sin ) / m

We can solve for or sin from

tan =^{opp}/_{adj}=^{14 cm}/_{200 cm}= 0.07

= 4.00°

sin = sin 4.00° = 0.07

d = 0.03 mm = 3 x 10^{ - 5}m

Therefore

= [(3 x 10^{ - 5}m)(0.07)] / 5

= 4.2 x 10^{ -7}m = 420 x 10^{ - 9}m = 420 nm

25.17 Red light ( = 640 nm) passes through a diffraction grating and on to a screen 2.0 meters away. There the first-order maximum or bright line appears 25 cm on either side of the central maximum. How many lines per centimeter are there on the grating?

First, determine the angle ,

tan =^{25 cm}/_{200 cm}= 0.125

= 7.125°

We expect to find maxima or bright regions for

d sin = m

Here we have only first-order maxima, with m = 1, on either side of the central maximum (m = 0) so we can solve for d, the separation distance between the "slits" of the diffraction grating,

d = m / sin = / sin

d = 640 nm / sin 7.125°

d = 640 nm / 0.124

d = 5160 nm = 5,160 x 10^{ - 9}m = 5.16 x 10^{ - 6}m

d = 5.16 x 10^{ - 4}cm

This is the slit separation distance. From this we can determine N the number of slits there are in a centimeter. If there were ten slits per centimeter (N = 10/cm) then the separation distance would be 0.1 cm (d = 0.1 cm). N and d are related by

N = 1/d

N = 1.94 x 10^{3}lines/cm = 1,940 lines/cm

25.28 When 634-nm light passes through a slit 0.1 mm wide and forms a diffraction pattern on a screen 1.25 m away, what is the distance from the central maximum to the center of the first minimum?

For a single-slit diffraction pattern, the first minimum occurs for

D sin =

or

sin = /D

sin = 634 nm/0.1mm

sin = ( 634 x 10^{ - 9}m ) /( 0.1 x 10^{ - 3}m)

sin = 6.34 x 10^{ - 3}= 0.00634

The distance y is given by

tan =^{y}/_{L}

where L is the distance from the slit to the screen, L = 1.25 m

y = L tan

For such small angles

tan sin

y = L sin

y = (1.25 m)(0.00634)

y = 7.93 mm

25.40 A high precision telescope advertises that it can resolve the eyes of a bird (assume them 1 cm apart) at 1.5 km. What is the minimum objective diameter?

The Rayleigh criterion for rsolution of two point sources is

= 1.22 / D

D = 1.22 /

tan = 1.0 cm / 1.5 km = 1.0 x 10^{ - 2}m / 1.5 x 10^{3}m

tan = 6.67 x 10 - 6

= 6.67 x 10^{ - 6}radians (= 3.8 x 10^{ - 4}degrees)

Since no particular wavelength is given for the light, use a wavelength near the middle of the visible spectrum such as

550 nm = 550 x 10^{ - 9}m

D = 1.22 /

D = 1.22 ( 550 x 10^{ - 9}m) / 6.67 x 10^{ - 6}

D = 10 cm

25.44 Magnesium chloride (MgCl_{2}), which has an index of refraction of 1.25, is often used for the non-reflecting coating on photographic lenses. How thick must a film of it be to have total destructive interference for 560 nm light?

For a thin film like this, with n_{1}< n_{2}and n_{2}< n_{3}, total destructive interference occurs for

2 t = [ m + (1/2)] ' = [ m + (^{1}/_{2})] / n

m is an integer. For the thinnest film, m = 0.

2 t = (^{1}/_{2})(560 nm) / 1.25

t = (^{1}/_{2})(560 nm) / [ (1.25) (2) ]

t = 112 nm

t = 1.12 x 10^{ - 7}m

25.62 When 580-nm light passes through a 1500 line/cm diffraction grating, how many orders of maxima do you see on a screen that is 1.5 m wide and1.2 m from the grating?

Recalling, from Problem 25.28, that d =^{1}/_{N}, we can determine the slit separation distance d from N, the number of lines/cm,

d = 6.67 x 10^{ - 6}m

Maxima for diffraction patterns occur for

d sin = m

The largest possible angle_{max}is

tan_{max}= 1.5 m / 1.2 m = 1.25

_{max}= 51.3°

sin_{max}= 0.78

Then the largest possible order m_{max}is given by

m = d sin /

m_{max}= d sin_{max }/

m_{max}= (6.67 x 10^{ - 6}m)(0.78) / (580 x 10^{ - 9}m)

m_{max}= 8.97

But m must be an integer! What does m_{max}= 8.97 mean? The maximum order of diffraction pattern (bright area) that we will see will be

m = 8

The next bright diffraction pattern, for m = 9, will be just off the edge of the screen.

25.63 An observer looks at a light source through a grating having 1000 lines/cm. If = 500 nm, how many images of the source can be seen? (Hint: For what order is = 90 degrees?)

Of course, this is similar to the previous problem. We can determine the slit separation distance d from N, the number of lines/cm,

d =1.0 x 10^{ - 5}m

Maxima for diffraction patterns occur for

d sin = m

The largest possible angle_{max}is 90° or sin_{max}= 1.0

m_{max}= d sin_{max}/

m_{max}= (1.0 x 10^{ - 5}m) (1.0) / (500 x 10^{ - 9}m)

m_{max}= 20

(c) Doug Davis, 1997; all rights reserved

Return to Ch 25, Wave Optics: Interference and Diffraction