33-42 A light ray in air strikes the right-angle prism shown in the figure (6 B = 28◦). This ray consists of two different wavelengths. When it emerges at face AB, it has been split into two different rays that diverge from each other by 8.50◦. (a) Find the index of refraction of the prism for each of the two wavelengths.
(a) In order to use Snell’s Law to find the index of refraction, it is necessary to determine the angle each of these light rays make with a line normal to the surface of the prism. This means working out a little geometry.
The angle θ between the incident ray and the normal to the hypotenuse of the prism is
θ = 90◦− 28◦= 62◦
Since the 12◦ angle is given, we can find the angle between the upper refracted ray and the normal:
θ1= 90◦− 28◦+ 12◦= 74◦.
The angle between the normal and the other refracted ray is θ2= θ1+ 8.5◦= 82.5◦
Now we can write Snell’s Law for these situations as nredsin 62◦= nairsin 74◦ and nbluesin 62◦= nairsin 82.5◦,
where nred (nblue) is the index for the red (blue) light. We aren’t told the exact colors, but we expect the light that is bent less (more) will be more to the red (blue) end of the spectrum. Now, letting nair= 1 be the index in air, we solve for each index:
33-45 Old photographic plates were made of glass with a light-sensitive emulsion on the front surface. This emulsion was somewhat transparent. When a bright point source is focused on the front of the plate, the developed photograph will show a halo around the image of the spot. If the glass plate has a thickness, t = 3.60 mm, and the halos have an
inner radius, R = 5.02 mm, what is the index of refraction of the glass? (Hint: Light from the spot on the front surface is scattered in all directions by the emulsion. Some of it is then totally reflected at the back surface of the plate and returns to the front surface.)
R/2 R/2 R
The figure above shows the rays that produce the the inner radius of the halo. A ray leaves from the point source at O, is totally internally reflected (at the critical angle θc) at A, and then reaches the inner radius of the halo at point B. We can relate θc to the index of refraction n of the glass using
sin θc= nair
n = 1 n.
Using the geometry shown in the picture we see that we can determine θc using
tan θc =R/2 refraction of the glass will be
n = 1 sin θc
= 1.75.
One can derive a simple expression for n by direct evaluation of sin θc using the sides of the right triangle ABC:
1
n= sin θc= R/2
√(R/2)2+ t2 =⇒ n =√
1 + (2t/R)2.
YF 34-57 A telescope is constructed from two lenses with focal lengths of 90.0 cm and 20.0 cm, and the 90.0 cm lens is used as the objective. Both the object being viewed and the final image are at infinity. (a) Find the angular magnifica-tion for the telescope. (b) Find the ABSOLUTE VALUE of the height of the image formed by the objective of a build-ing 60.0 m tall, 3.00 km away. (c) What is the ABSOLUTE VALUE of the angular size of the final image as viewed by an eye very close to the eyepiece? Give your answer in RA-DIANS.
December 8, 2011
34-89 Two thin lenses with focal lengths of magnitude 15.0 cm, the first diverging and the second converging, are placed 12.00 cm apart. An object 5.00 mm tall is placed 5.00 cm to the left of the first (diverging) lens. (a) Where is the image formed by the first lens located? (b) How far from the object is the final image formed? (c) Is the final image real or virtual? (d) What is the height of the final image?
(e) Is the final image erect or inverted?
(a) In a two lens system, the image from the first lens, serves as the object of the second lens. The location of the image formed by the first lens can be found using do = 5 cm, f =
−15 cm, and solving the lens equation for di. 1 to the left of the lens, on the same side as object. It is an upright, virtual image. We know that it is upright because the magnification is a positive number. We know that it is virtual because di is a negative number. A ray diagram confirms this.
(b) To find the distance from the final image to the object, we need to determine the location of the final image. The image from the first lens becomes the object for the second lens. We need to determine do: because the image from the first lens is 3.75 cm to the left of the first lens, that means it is 12 cm + 3.75 cm = 15.75 cm from the second lens. Using do = 15.75 cm and f = +15 cm, solve for di = 315 cm.
This means that the final image is 315 cm to the right of the second lens. So the distance between the object and final image is 5 cm + 12 cm + 315 cm = 332 cm.
(c) The final image is real, because we get a positive image distance (di) for the second lens, and we can tell from the ray diagram.
(d) We can find the height of the final image by multiplying the magnification of each lens.
hi= m1m2ho= Mastering Physics wants height (absolute value of hi) in units of cm, so enter 7.5.
(e) The final image is inverted, because the total magnifica-tion is negative, and the ray diagram confirms this.
34-91 An object to the left of a lens is imaged by the lens on a screen 30.0 cm to the right of the lens. When the lens is moved 4.00 cm to the right, the screen must be moved 4.00 cm to the left to refocus the image. Determine the focal length of the lens.
For the first set up, the image distance is di = 30.0 cm and let the object distance be the variable do.
Thus, the equation for the focal length is 1
In the second set up the object distance is increased by 4.00 cm, thus d0o = do+ 4.00 cm. The image distance is decreased by 8.00 cm (4.00 cm from the lens moving to the right and another 4.00 cm for the screen moving to the left).
Therefore the new image distance is d0i = di− 8.00 cm = 30.0 cm− 8.00 cm = 22.0 cm.
These values can be used in another equation for the focal length We can set the expressions for 1/f given by Eqs. (1) and (2) equal to each other and use the quadratic equation to solve for do:
The quadratic equation for do has two roots, but do must be positive because the object is on the side of the lens from which the light comes. Therefore we used the plus sign in the quadratic formula to ensure do > 0. Now do and the original di can be used to solve for the focal length f :
1
35-8Young’s experiment is performed with light from ex-cited helium atoms (λ = 502 nm). Fringes are measured carefully on a screen 1.20 m away from the double slit, and the center of the twentieth fringe (not counting the central bright fringe) is found to be 10.6 mm from the center of the central bright fringe. What is the separation of the two slits?
35-12Coherent light with wavelength 400 nm passes through two very narrow slits that are separated by d = 0.200 mm and the interference pattern is observed on a screen L = 4.00 m from the slits. (a) What is the width of the central interference maximum? (b) What is the width of the first-order bright fringe?
(a) The position x = x0 of the first dark spot is given in terms of the angle it makes with with central bright spot (at x = 0) by
d sin θ≈ dx0
L = (m +12)λ =⇒ x0= Lλ 2d,
where we invoked small-angle approximations and set m = 0 since we are interested in the central peak. We have
x0=(4.00 m)(400× 10−9m)
2(2× 10−4m) = 0.004 m = 4 mm The width of the central interference maximum will be dou-ble the distance from the bright central spot to the first dark spot, which is 2x0 = 8.0 mm.
(b) The width of the first order bright fringe will be the distance between the first and second dark spots. We already know the position of the first dark spot. All we must do now is calculate the position of the next dark spot (at x1, for m = 1) and find the difference (x1− x0):
d sin θ≈ dx1
L = (1 +12)λ =⇒ x1= 3Lλ 2d ,
Clearly x1 = 3x0 = 12 mm, so the width of the first order bright fringe is
x1− x0= 12 mm− 4 mm = 8 mm.