course Phy232
Assignment 18 Ch. 34 Problems 62, 71, 79, 86, 91, 98, 10571. Longitudinal magnification is defined as m=ds/ds
(A): Show that for a spherical mirror. M=-m2. What is significance of the fact that m is always negative?
A negative value denotes an inverted image. Longitudinal magnification denotes the factor by which an image increases in size.
(B): Concave mirror with radius of curvature=150.0cm. Cube faced towards mirror is 200.0cm to left of mirror vertex.
(i): Find the location of the image of this face and opposite face.
120.00cm from mirror, 119.96cm from mirrot
(ii): m=-0.600, m’=-0.390
(iii): faces perpendicular to the axis, squares with side 0.600 mm
you need to show the details of your calculations
79. What should be the index of refraction of a transparent sphere in order for paraxial rays from an infinitely distant object to be brought to a focus at the vertex of the surface opposite the point of incidence?
2.00cm
An index of refraction is a pure number, not measured in cm.
2.00 is a reasonable and perhaps correct answer, but you need to show the details of your analysis.
86.
(A): Prove that na/nb=f/f
S is the object distance. S is the image distance for a spherical reflector we know that na/s+nb/s= na-nb/R
If “s” is infinite then na/s is 0 an imagine distance s’=f’ so nb/i=nb/f’= (nb-na)/R
Nb/f’=na/f which is rearranged to get na/nb=f/f
(B): f/s+f’/s’=1
Nb/f’=(nb-na)/R we get f’=nbx R/(nb-na)
Na/f=(nb-na)/R we get f=na x R/ (nb-na)
Rearranging na/s=nb/s’=(nb-na)/R we get Rxna/(s9na-nb))+Rxnb/(s’(na-nb))=1
Combining we get (f/s+f’/s)/1
91. Three lenses each with focal length 40.0cm are aligned on a common axis. Adjacent lenses are separated by 52.0cm. Find the position of the image of a small object on the axis, 80.0cm to the left of first lens.
F-#=Focal length/diameter (f/D)
= 40/52 + 40/104 + 40/156
= 1.41
Each image is the object for the next lens. You need to find the position of the image of the first lens, which becomes the object for the second. Then find the position of the image of this object through the second lens. Repeat the process for the third lens.
98.
(A): Equation (34.19) assumes that the lens is surrounded by air. Consider a thin lens immersed in a liquid with refractive index nliq. Prove that the focal length f’ is given by Eq (34.19) with n replaced by n/nliq.
1/f=(n-1)(1/R1-1/R2)
Well n is the refraction of air which in this problem we a surrounded by a liquid so you would have to divide by the liquid.
(B): A thin lens with index n has focal length “f” in vacuum.
F’=[nliq(n-1)/n-nliq]
1/f=(n-1)(1/R1-1/R2)
1/f= nliq/n(n-1_
F=nliq(n-1)/n-nliq
Your solution should start by modifying the assumption of the original derivation, that the index of refraction of the surrounding material is air with an index of refraction of 1. If you follow the steps of the derivation replacing this number by n_liq, you will get the desired result.
105.
(A): For a distant object for which the image distance is approximately equal to the focal length of the camera lens. Show that the angle of view theta is given by theta=2arctan (d/2t)
(B):
Theta=2tan(28/2f)= 75 degrees
Theta= 2tan(105/2f)= 23 degrees
Theta= 2tan (300/2f)= 8.2 degrees
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