TOTAL INTERNAL REFLECTION AND
LENSES
GROUP 4
Firdiana Sanjaya (4201414050)
Ana Alina (4201414095)
• What happened? Why did light not refract through
the second face?
• The phenomenon observed in this part of the lab is
known as total internal reflection. Total internal
reflection
• Total internal reflection is a phenomenon which occurs
when a propagating wave strikes a medium boundary at
an angle larger than a particular critical angle with respect
to the normal to the surface. The critical angle is
the angle of incidence above which the total internal
reflection occurs.
• The critical angle can be found from Snell's law, putting in
an angle of 90° for the angle of the refracted ray. This
gives:
• For any angle of incidence larger than the critical angle,
Snell's law will not be able to be solved for the angle of
refraction, because it will show that the refracted angle
has a sine larger than 1, which is not possible.
• Total internal reflection will not take place unless the incident light is
traveling within the more optically dense medium towards the
less optically dense medium. TIR will happen for light traveling from
water towards air, but it will not happen for light traveling from air
towards water. TIR occurs because the angle of refraction reaches a
90-degree angle before the angle of incidence reaches a 90-degree
angle. The only way for the angle of refraction to be greater than the
angle of incidence is for light to bend away from the normal. Since light
only bends away from the normal when passing from a more dense
medium into a less dense medium, then this would be a necessary
condition for total internal reflection.
Depth perception
• For any angle of incidence larger than the critical angle,
Snell's law will not be able to be solved for the angle of
refraction, because it will show that the refracted angle
has a sine larger than 1, which is not possible.
Example
• A beam of light travels from water into a piece of diamond in the
shape of a triangle, as shown in the diagram. Step-by-step, follow the
beam until it emerges from the piece of diamond.
(a) How fast is the light traveling inside the piece of diamond?
The speed can be calculated from the index of refraction:
(b) What is , the angle between the normal and the beam of
light inside the diamond at the water-diamond interface?
• A diagram helps for this. In fact, let's look at the complete
diagram of the whole path, and use this for the rest of the
questions.
The angle we need can be found from Snell's law:
• (c) The beam travels up to the air-diamond interface. What is , the
angle between the normal and the beam of light inside the diamond at
the air-diamond interface?
• This is found using a bit of geometry. All you need to know is that the
sum of the three angles inside a triangle is 180°. If is 24.9°, this
means that the third angle in that triangle must be 25.1°. So:
• (d) What is the critical angle for the diamond-air interface?
• (e) What happens to the light at the diamond-air interface?
• Because the angle of incidence (64.9°) is larger than the critical
angle, the light is totally reflected internally.
• (f) The light is reflected off the interface, obeying the law of reflection.
It then strikes the diamond-water interface. What happens to it here?
• Again, the place to start is by determining the angle of incidence, . A
little geometry shows that:
• The critical angle at this interface is :
• Because the angle of incidence is less than the critical angle, the
beam will escape from the piece of diamond here. The angle of
refraction can be found from Snell's law:
Lenses
• A lens is merely a carefully ground or molded piece of transparent
material that refracts light rays in such as way as to form an image.
Lenses can be thought of as a series of tiny refracting prisms, each of
which refracts light to produce their own image. When these prisms
act together, they produce a bright image focused at a point.
• We shall consider only the special case of a thin lens—that is, a lens
in which the thickest part is thin relative to the object distance p, the
image distance i, and the radii of curvature r1 and r2 of the two
surfaces of the lens. We shall also consider only light rays that make
small angles with the central axis (they are exaggerated in the figures
here).We shall prove that for such rays,a thin lens has a focal length
f.Moreover, i and p are related to each other by
• which is the same as we had for mirrors.We shall also prove that
when a thin lens with index of refraction n is surrounded by air, this
focal length f is given by
• Magnification of Images
• Consider a thin lens through which light rays from an
object pass. As with mirrors, we could analyze a
geometric construction to show that the lateral
magnification of the image is
• From this expression, it follows that when M is positive,
the image is upright and on the same side of the lens as
the object. When M is negative, the image is inverted and
on the side of the lens opposite the object.
Convex lens
(a) When the object is in front of and outside the focal point of a
converging lens, the image is real, inverted, and on the back side of the
lens. (b) When the object is between the focal point and a converging
lens, the image is virtual, upright, larger than the object, and on the
front side of the lens.
Concave lens
• When an object is anywhere in front of a diverging lens,
the image is virtual, upright, smaller than the object, and
on the front side of the lens.
Sign Convention
• Because a lens transmits light rather than reflecting it like a
mirror does, the other side of the lens is the positive side for
images. In other words, if the image is on the far side of the
lens as the object, the image distance is positive and the image
is real. If the image and object are on the same side of the
lens, the image distance is negative and the image is virtual.
• For converging mirrors, the focal length is positive.
• A positive magnification corresponds to an upright image, while
a negative magnification corresponds to an inverted image
• Certain cases involving more than one lens the object distance
can be negative. This occurs when the image from the first lens
lies on the far side of the second lens; that image is the object
for the second lens, and is called a virtual object.
Example