Question

5.) Suppose that a point source is emitting waves uniformly in all directions. If a stationary observer A is twice as far from the same point source than stationary observer B, then a.) A will detect a higher intensity than B. b.) A and B will detect the same intensity. c.) A will hear a lower intensity than B. Answer __________ 6.) A light ray traveling in water, n = 1.3, is incident (angle of incidence = 45 degrees) upon a boundary with another medium, n = 1.2, therefore, the light ray transmitted into the second medium is refracted as follows; a.) transmission angle is less than 45 degrees b.) transmission angle is equal to 45 degrees c.) transmission angle is greater than 45 degrees d.) transmission angle is unknowable Answer __________ 7.) The mirror equation Do-1 + Di-1 = F-1 in which Do is the object distance, Di is the image distance and F is the focal length relates the three variables Do , Di , and F for any given mirror focusing an image of an object. The exponent (-1) on each of the variables means calculate the multiplicative inverse of the variable. As a numerical example, consider 4-1. 4-1 = ¼ = 0.25 Given Do = 10.0 cm, then determine the value of Do-1 = a.) 0.10 cm-1 b.) 0.20 cm-1 c.) 0.40 cm-1 d.) 0.80 cm-1 Answer__________ 8.) An object is placed 3.5 meters in front of a plane mirror. Determine the image distance. a.) - 6.5 meters b.) - 4.5 meters c.) - 3.5 meters d.) - 1.5 meters Answer __________

Answer 1

5)

Intensity = power/(area) = P/(4*pi*r^2)

IA/IB = (rB/rA)^2

given distance of A = rA = 2*distance of B rB

rA = 2rB

IA = IB*(rB/2rB)^2

IA = IB/4

IA < IB

Answer : A will hear a lower intensity than B

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6)

from snell law

nwater*sini = n*sinr

1.3*sin45 = 1.2*sinr

r = 50

Answer : c.) transmission angle is greater than 45 degrees

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7)

Do = 10 cm

Do-1 = 1/10 = 0.10 cm-1

ANSWER ( a)

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8)

for plane mirror

image distance di = -d0

image distance di = -3/5 meters

ANSWER ( c)