Geometric Optics Example Problems with Solutions PDF

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Geometric Optics Example Problems with Solutions...

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Geometric Optics: Example Problems with Solutions The Law of Refraction 1. Calculate the index of refraction for a medium in which the speed of light is 2.012 x 103 m/s. Solution 3  10 8 m/s c n n  1.46 n v 2.012  108 m/s 2. A coin is placed at a depth of 15 cm in a beaker containing water. The refractive index of water is 4/3. Hint: for small angles tan θ = sin θ. a. Calculate the apparent depth of the coin. b. By how much is the image of the coin raised? y θ2

θ1 Real depth d

Apparent depth da

θ2

Solution a. You can see the coin because of the light reflected of its surface. The light ray originates from the surface of the coin and travels through water (medium 1) and when it meets the water-air boundary (air = medium 2) at any angle other than 0 it will refract away from the normal. However the virtual image of the coin appears along the extension of the bent ray (in red), so you see a virtual image of the coin at a smaller depth, the apparent depth. y tan θ 1 d da y y tan θ1  tan θ 2    d  0.15 m tan θ 2 y d d da d n 1 4 /3 a S nell's Law : n 1sin θ 1  n 2sin θ 2 n 2  1 (air)

da ?

sin θ 1 n2 tan θ 1   sin θ 2 n1 tan θ 2



n2 d a  n1 d

 da 

d a  11.25 m b.

Δd  d - da

Δd  15 cm - 11.25 m

Δd  3.75cm 1

n2 d n1

da 

1 15 cm  4   3

3. Suppose you have an unknown clear substance immersed in water, and you wish to identify it by finding its index of refraction. You arrange to have a beam of light enter it at an angle of 45.00, and you observe the angle of refraction to be 40.30. What is the index of refraction of the substance and its likely identity? The index of refraction of water is 1.33. Solution S nell's Law : n 1sin θ1  n 2 sin θ 2 θ1  45.0 0 θ1 sin θ1 sin 45.00 n1 θ 2  40.3 0 n 2  n1 n 2  1.33 0 sin θ 2 sin 40.3 n 1  1.33 n 2  1.45 n2 n2 ? θ 2

From Table 25.1, the most likely solid substance is fused quartz.

Total Internal Reflection 4. A ray of light, emitted beneath the surface of an unknown liquid with air above it, undergoes total internal reflection as shown in figure. What is the index of refraction for the liquid and its likely identification? θ2 Solution φ Second medium is air:

n2  1 (air) S nell's Law : n1 sin θ1  n2 sin θ2 For θ c : n 1 sin θc  n 2 sin 90 0 n1 sin θc  n 2  1 n1 

sin 90 0  1

1 sin θ c

13.4 cm  θc  tan-1    15.0 cm  1 θc  41.78 0 n1 n 1  1.5 sin 41.780 Image Formation by Lenses tan θ c 

13.4 cm 15.0 cm

From Table 25.1: Benzene

5. What is the focal length of 1.75 D reading glasses found on the rack in a pharmacy? Solution The Power of a lens P has the unit of diopters (D p. 1003 text book): 1 1 1 P f  f f  0.571 m f  57.1 cm f P 1.75 D

2

6. How far from the lens must the film in a camera be, if the lens has a 35.0 mm focal length and is being used to photograph a flower 75.0 cm away? Solution f  35 mm

d o  750 mm di  ? 1 1 1   do di f 1 1  d i    f di d i  36.7 mm

 1

  

1 1 1  di f di 1 1  d i     35 mm 750 mm 

1

7. A doctor examines a mole with a 15.0 cm focal length magnifying glass held 13.5 cm from the mole. (a) Where is the image? (b) What is its magnification? (c) How big is the image of a 5.00 mm diameter mole? 1 1 1 1 1 1 a.     Solution d o di f di f d i 1 f  15 cm 1 1 1  1  1   di    d i    d o  13.5 cm  15 cm 13.5 cm   f di  a. d i  ? d i  - 135 cm di 1 The image is larger than the object.

c. m 

hi ho

h i  m h o  10 5.00 mm 

h i  5 cm

o

F

F

di do

3

8. An object 1.50 cm high is held 3.00 cm from a person’s cornea, and its reflected image is measured to be 0.167 cm high. c. What is the magnification? d. Where is the image? e. Find the radius of curvature of the convex mirror formed by the cornea. (Note that this technique is used by optometrists to measure the curvature of the cornea for contact lens fitting. The instrument used is called a keratometer, or curve measurer.) Solution

a. m 

hi ho

b. m  -

c.

m

0.167 cm 1.50 cm

di di  - m d0 do

1 1 1   do di f f  0.376 cm



m  0.11

d i  - 0.11 3 cm

1 1 1   f 3 cm 0.334 cm  R  - 2  0.376 cm

d i  - 0.334 cm behind the cornea

R  - 0.752 cm

object

image F

4

C...