Reflection and Refraction and Mirrors and Lenses

1. A light ray is moving from air into a piece of glass with an Index of Refraction = 1.60. The light

ray makes an angle of incidence of 30o, as shown in the diagram below.

Glass n = 1.60

Θi = 30o

Θrefl = 30o

Θ2

8 m/a. What is the speed of light in the glass? The speed of light in a vacuum is c = 3.0*10 s?

883.00 10 m 1.875 10

1.60 sc cn v vv n

×= = = = ×

b. Determine the angle of reflection and the angle of refraction.

( ) ( )( ) ( )

( )

( ) o1-2

o

2

2o

2211

18.2 0.3125 Sin

0.3125 601500

60130Sin 1.00 Sin

Sin 1.60 30Sin 1.00

Sin n Sin n

==Θ

==⋅

=Θ

Θ⋅=⋅

Θ⋅=Θ⋅

.

..

c. A second ray is moving from the Flint Glass into air and makes an angle of incidence of 45o.

. A light ray is passing from air into water.

. If the index of refraction for air is n = 1 and the incident angle is 30°, what is the angle of the

Determine the angle of refraction and sketch the path of the light ray after it hits the boundary.

Snell’s Law:

Glass n = 1.60

Θi = 45o

Θr = 45o

( ) ( )( ) ( )

( )

( ) !impossible 1.13 Sin

1.13 001131

00145Sin 1.60 Sin

Sin 1.00 45Sin 1.60

Sin n Sin n

1-2

o

2

2o

2211

,..

.=Θ

==⋅

=Θ

Θ⋅=⋅

Θ⋅=Θ⋅

2

a

refracted beam?

The index of refraction for water is n = 1.33. By Snell's Law, 2211 sinsin inin = ,

the angle of the refracted beam is:

( )1 11 1sin sin 22.1i = = = °

b. What is the angle the beam makes with the normal as it exits the water?

In this case, the roles of n1 and n2 are interchanged. Then Snell's Law yields:

22 1.33n

sin 30sinn i− − °

( ) ( )1 12 2 1.33 sin 22.1sinn i− − °1

1 1nsin sin 30.0i = = = °

3. An object is placed 20 cm from

. a convex lens, of focal length 15 cm.

cm.

ication in each case.

a

b. a concave lens, of focal length 15

Calculate the image position and magnif

a. Plug in s0 and f into the formula, fss io

11= , where s0 = +20cm and f = +15 cm, We obtain

s magnification

1+

1s1 = +60 cm. The image is real and ha 0

3.020

ms

= = = .

b. Plug in s

60s

fss111

io

=+ , whe f =0 and f into the formula, re s0 = +20cm and -15 cm, we obtain

s magnificatio 1

0

0.4320

ms

= . s1 = -8.6 cm. The image is virtual and ha n 8.6s

= =

4. Consider a converging lens that has a focal length of f = 10 cm. An object of height 1 cm is

laced in front of the lens at 5 cm. Characterize the resulting image. The thin lens equation: p

fss111

io

=+

Since the lens is converging, the sign of f is positive. s0 =+5 cm since the image is on the object

side of the lens. We can then find s1': fss111

=+ io

1 0 1 1

1 1 1 1 1 1 1 1 2 110 5 0s f s s s 10 1

−= − = − = = − .

So, s1 = -10 cm, which means that the image is on the object side of the lens. The magnification

of the image is given by M=-s1/s0 = -(-10/5)=+2. The height of the image is +2 cm.

• The image is upright since m>0.

• The image is enlarged since |m|>1.

• The image is virtual since s < 0. 1