Chapter 35: Reflection Section 16: Mirror equations 58 cm, ?2.4 200 cm, 2.4 200
ID: 1866120 • Letter: C
Question
Chapter 35: Reflection
Section 16: Mirror equations
58 cm, ?2.4
200 cm, 2.4
200 cm, ?2.4
58 cm, 0.41
58 cm, ?0.41
(0/2 submissions used)
Chapter 36: Refraction
Section 3: Snell’s law
(0/5 submissions used) °
Chapter 40: Special Relativity
Section 12: Length contraction
Section 16: Mass and energy
What are the corrects answers and how did you solve for them?
AP35.16.1(5.00) A concave spherical mirror with radius of curvature 82 cm forms an image of an object located 140 cm in front of it. The distance of the image from the mirror and the mirror’s magnification are, respectively, most nearly
58 cm, ?2.4
200 cm, 2.4
200 cm, ?2.4
58 cm, 0.41
58 cm, ?0.41
(0/2 submissions used)
Explanation / Answer
Chapter 35: Reflection
Section 16: Mirror equations
Ans: 58 cm, ?0.41
focal length of the mirror, f = R/2 = 82/1 = 41 cm
object distance distance, u = 140 cm
let v is the image distance.
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/41 - 1/140
==> v = 58 cm
magnification, m = -v/u = -58/140 = -0.41
Chapter 36: Refraction
Section 3: Snell’s law
Use Snell's law, sin(theta_i)/sin(theta_r) = n2/n1
sin(79.3)/sin(theta_r) = 1.52/1.11
==> sin(theta_r) = sin(79.3)*1.11/1.52
theta_r = sin^-1(0.718)
= 45 degrees
Chapter 40: Special Relativity
Section 12: Length contraction
use Length contraction equation,
L = Lo*sqrt(1 - (v/c)^2)
= 369*sqrt(1 - (2.7*10^8/(3*10^8))^2)
= 161 m
Section 16: Mass and energy
using Einstein's mass-enery equivalence equation,
E = m*c^2
= 38*10^-3*(3*10^8)^2
= 3.42*10^15 J
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