Geometrical optics, or ray optics, is a model of optics that describes light propagation in terms of rays. The ray in geometric optics is an abstraction useful for approximating the paths along which light propagates under certain circumstances.
The simplifying assumptions of geometrical optics include that light rays:
propagate in straight-line paths as they travel in a homogeneous medium
bend, and in particular circumstances may split in two, at the interface between two dissimilar media
follow curved paths in a medium in which the refractive index changes
may be absorbed or reflected.Geometrical optics does not account for certain optical effects such as diffraction and interference. This simplification is useful in practice; it is an excellent approximation when the wavelength is small compared to the size of structures with which the light interacts. The techniques are particularly useful in describing geometrical aspects of imaging, including optical aberrations.
The geometric configuration that I am adopting is the following, I hope you understand.
The optical fiber is positioned relative to the bottom surface at a height ##a## and an angle ##\alpha## with respect to the y-axis in the yz-plane with x = 0. ##b## is the distance between the origin and...
I would know how to solve this problem if the person had been standing pratically above of the object underwater by using Snell's law and the approximation ##\sin(\theta)\approx\tan(\theta)## fopr ##\theta## small, but in this case I don't see how to find the angles ##\theta_1## and ##\theta_2##...
Using the data given and recalling that in this configuration ##R<0## I get: ##\frac{1.33}{0.5}+\frac{1.5}{q}=\frac{1.5-1.33}{-0.2}\Rightarrow q\approx -0.427 m=-42.7 cm## so the image is virtual and is ##42.7\ cm## to the left of vertex ##V##. The magnification is ##M=\frac{n_1 q}{n_2...
I've heard virtual objects are used in cinema halls I'm not sure of what that means, can someone explain this to me please?
And btw by virtual object i mean when diverging rays from real object passes through a convex lens, the rays will converging and if you'll place a plane mirror infront of...
if someone want to explain to me what is an upright image ? , and what are the other adjectives to define an image in geometric optics and their meaning , Thanks .
From ray tracing I would say that the image is upright.
Using the equation ##\frac{1}{p}+\frac{1}{q}=\frac{1}{f}## with ##f=-\frac{R}{2}=-2## and ##M=-\frac{q}{p}=\frac{3}{4}## I get ##p=\frac{2}{3}cm\simeq 0.67 cm##.
Is this correct? Thanks
I tried using the formula for the refraction of a spherical lens ##\frac{n_1}{p}+\frac{n_2}{q}=\frac{n_2-n_1}{R}## consider each slab as a spherical lens with curvature ##R=\infty## and by doing that I get ##\frac{1.33}{10}+\frac{1.5}{q}=0\Leftrightarrow q\approx -11.3 cm##. Since the piece of...
I used the equation for the refraction on a spherical surface: ##\frac{n_1}{p}+\frac{n_2}{q}=\frac{n_2-n_1}{R}##, where ##n_1=1## is the index of refraction of air, ##n_2## the index of refraction of the sphere, ##R## is the radius of the glass sphere, ##p## is the object distance which, since...
I created the following ray diagram to help me solve the problem:
Then I applied the mirror equation 3 separate times.
However, the final image distance I got is wrong. I'm wondering if I'm mistaken in taking the last object distance to be negative. However I only have one more try to get this...
From This picture, I think the fish will be smaller but the problem is how small will it be?
(Fish "L" is the image of fish "K")
Let ##H## be the depth of fish "K", ##\theta## be the angle of eyes to y-axis and ##n## is the index of refraction of water.
Homework Statement:: Finding the distance between the back surface of the first lens and the front surface of the back lens.
Homework Equations:: 1/f = 1/s_o + 1/s_i
I have two positive thin lenses that are separated by a distance of 5 cm. The focal lengths of the lenses are F_1 = 10 cm and...
I've tried to attempt the first part of the problem(spent over an hour on this) as second part could be easily optained with some calculus ,I asked my friend but alas nobody could conjure the solution to this dangerous trigonometric spell.
It was just pages and pages of concoction of...
I am extremely confused by the use of the term coefficient of increase of something. For example , if it is stated that the index of refraction varies linearly with a coefficient of 2.5×10^-5, how is this coefficient defined? Is it simply the slope of the line plotted with index of refraction on...
Homework Statement
Derive an analytic expression for the distance from the vertex to the focus for a particular ray in terms of (i) the radius of curvature R of the concave mirror (ii) the angle of incidence θ between incident ray and radius of the mirror. Hence show that the focus moves closer...
Is there a simple way to determine or prove this? Real images are always inverted, and unlike virtual images can be projected onto a screen (I'm not even sure what this means to be honest). If I look at the back end of a spoon (convex mirror), the image is always upright and therefore virtual...
Homework Statement
At the outside, there is a vertical stick with a length of 1.1 meter and its shadow on the surface of an Earth is 1.3 meter, there also is light pole and its shadow length is 5.2 meters, what is the height of that light pole?
Homework Equations
Trigonometry equations to...
The focal of the lens equivalent of two thin lens at distance h is
$$1/f=1/f_1+1/f_2+h/(f_1 f_2)$$
Therefore, supposing that ##f_1>0## and ##f_2>0## (both lenses are convergent), if ##f_1+f_2 <h## then the equivalent lens should be divergent.
Nevertheless consider the example in picture...
Homework Statement
Draw the ray diagram of the case of a diverging lens where both object and image are virtual, that is ##f<0## , ##p<0## with ##p<f##
Homework Equations
The Attempt at a Solution
I did not find this particular case of ray diagram in any textbook so I would like to know...
Homework Statement
Homework Equations
We will call ##s## the distance of the object from the first lens, ##s'## the distance of the image from the first lens and ##s''## the distance of the image from the second lens.
The Gauss's lens equation:
$$\frac{1}{s} +\frac{1}{s'}=\frac{1}{f_1}$$...
Homework Statement
A person's eye has a near point of 7 cm. The cornea at the outer surface of the eye has a refractive index of n_c = 1.376 and forms a convex shape with a radius of curvature of R_2 = 8 mm against air. The figure below shows the same eye with a contact lens (refractive index...
A plane mirror forms a virtual image of a real object placed in front of it and a real image of a virtual object placed in front of it. I can't picture the second case. Please show me a ray diagram showing real image formation by a plane mirror or just explain the case of real image formation by...
Homework Statement
Hi!
I have lately come across some tricky experimental physics tasks, where no solution is given. Some of them involved parts where the focal length of a convex lens had to be measured with a laser. How do you do this?
Homework Equations
1/l + 1/d = 1/f (1)
where f is the...
I am trying to follow the derivation of lens maker's formula from the textbook "University Physics", p.1133 (https://books.google.com.hk/books?id=nQZyAgAAQBAJ&pg=PA1133#v=onepage&q&f=false [Broken])
I can understand the first equation because it is just the object–image relationship for...
Homework Statement
Given a "new type" of optical fiber (index of refraction n = 1.23), a laser beam is incident on the flat end of a straight fiber in air. Assume nair = 1.00. What is the maximum angle of incidence Ø1 if the beam is not to escape from the fiber? (See attached file for...
1.Problem: A vessel is quarter filled with a liquid of refractive index A. the remaining parts of the vessel is filled with an immiscible liquid of refractive index 3A/2. The apparent depth of the vessel is 50% of the actual depth. the value of A is?
1)1 2)3/2 3)2/3 4)4/3
Solution: 1...
Hi,
I am familiar with drawing rays through a lens. But when a few lenses are put together, things become confusing to me.
For example, if a first positive thin lens at 0 forms a real image 10cm away, what would happen when we put a second positive thin lens, say at 5cm along the optical...