Physics 112 – Berry College – Spring 2014 1 Instructor: Usama al-­‐Binni Recitation Session 02 Problem 1 Suppose you’d like to design a full-­‐length, silvered-­‐glass mirror that a person who is anywhere from 5'4" to 6'5" can use. Assume that a person’s eyes are on average 14.0 cm from the top of their head. 1. How high must the top of the mirror be for a 6'5" person to be able to see the top of their head in the mirror? Can a 5'4" person also see the top of their head in this case? 6! 5!! = 6×12 + 5 = 7!! = 77!! ×
196 cm −
2.54 cm
= 196 cm 1 inch
14.0 cm
= 189 cm 2
This would be enough for the 5'4" person as well 2. How low must the bottom of the mirror be for a 5'4" person to see his feet? Can a 6'5" guy also see their feet in this case? 5! 4!! = 5.33 ft = 5.33 ft
!" !"
! !"
×
!.!" !"
! !"
= 163 cm 163 cm − 14 cm = 149 cm 149
cm = 75 cm above ground 2
The 6′5′′ would still be able to see their feet, since the minimum height in their case is: 196 cm − 14 cm
= 91 cm 2
3. How tall must the mirror be to accommodate people from 5'4" to 6'5" in height? 189 cm − 75 cm = 114 cm Physics 112 – Berry College – Spring 2014 2 Instructor: Usama al-­‐Binni Problem 2 A glass (𝑛 = 1.52) cup is filled with oil (𝑛 = 1.44) and an object is placed inside. Consider a ray of light that travels from the object, through the oil, and into the glass. 1. Can the ray undergo total internal reflection when it hits the inner surface of the glass? Since 𝑛! = 𝑛!"# is less than 𝑛! = 𝑛!"#$$ we cannot reach a critical angle and no total internal reflection occurs. 2. What is the critical angle for total internal reflection when the ray strikes the exterior surface of the glass? Assume that there is air outside of the glass. 𝑛! sin 𝜃!,! = 𝑛! sin 𝜃! ⇒ 𝜃!,! = sin!!
𝑛!
1.00029 sin 90°
sin 𝜃! = sin!!
= 41.2° 𝑛!
1.52
3. What must the angle of incidence be on the interior surface of the glass for the ray to strike the exterior surface of the glass at this critical angle? 𝑛! sin 𝜃! = 𝑛! sin 𝜃! 𝜃! = sin!!
𝑛! sin 𝜃!
1.52 sin 41.2°
= sin!!
= 44.0° 𝑛!
1.44
4. Calculate the critical angle for total internal reflection if there were no glass present (in other words, if the light hits a boundary between the oil and air). Does the presence of the glass have any effect on whether or not any light can make it out into the air? 𝑛! sin 𝜃!,! = 𝑛! sin 𝜃! ⇒ 𝜃!,! = sin!!
𝑛!
1.00029 sin 90°
sin 𝜃! = sin!!
= 44.0° 𝑛!
1.44
Physics 112 – Berry College – Spring 2014 3 Instructor: Usama al-­‐Binni Problem 3 Problem 5 on p. 774 from the book Two identical plane mirrors of width 𝑤 are placed a distance 𝑑 apart with their mirrored surfaces parallel and facing each other. a. A beam of light is incident at one end of one mirror so that the light just strikes the far end of the other mirror after reflection. Will the angle of incidence be: a. sin!! (𝑤/𝑑) b. cos !! (𝑤/𝑑) c. tan!! (𝑤/𝑑) tan 𝜃 =
𝑤
⇒ 𝜃 = tan!! (𝑤/𝑑) 𝑑
b. If 𝑑 = 50 cm and 𝑤 = 25 cm, what is the angle of incidence? 𝜃 = tan!! (𝑤/𝑑) = tan!!
1
= 26.6° 2
Physics 112 – Berry College – Spring 2014 4 Instructor: Usama al-­‐Binni Problem 4 An object with a height of 30.0 cm is placed 1.00 m from a convex spherical mirror. The spherical mirror has a radius of curvature of 𝑅 = 42.0 cm. 𝑓=
𝑅
1
1
1
𝑑!
ℎ!
, =
+ , 𝑚 = −
= 2
𝑓 𝑑! 𝑑!
𝑑! ℎ!
a. What is the focal length of the mirror? 𝑓=
𝑅 −42.0
=
= −21.0 cm 2
2
b. Use the mirror equation to calculate the distance of the image from the mirror. 1
1
1
1
1
1
=
+ ⇒
=
+ ⇒ 𝑑! = −17.4 cm 𝑓 𝑑! 𝑑!
−0.210 1.00 𝑑!
c. Find the magnification and the height of the image. 𝑚=−
𝑚=
𝑑!
−17.4
=−
= 0.174 𝑑!
100
ℎ!
ℎ!
= 0.174 =
⇒ ℎ! = 5.22 cm ℎ!
30.0
d. Is the image real or virtual? Is the image upright or inverted? Since 𝑚 > 0, the image is upright, and since 𝑚 < 1, it is reduced in size. The image is virtual, since, 𝑑! < 0.