Consider a point object O in front of a concave refracting...

Consider a point object $O$ in front of a concave refracting surface $S P M$ separating two media of refracting indices $n_{1}$ and $n_{2}$ (see Fig. 3.34); $C$ represents the center of curvature. In this case also one obtains a virtual image. Let $Q$ represent an arbitrary point on the axis. We now have to consider the optical path length $L_{\mathrm{op}}=n_{1} O S-n_{2} S Q$; show that it is given by $$ \begin{aligned} L_{\mathrm{op}} &=n_{1} O S-n_{2} S Q \\ & \approx n_{1} x-n_{2} y-\frac{1}{2} r^{2}\left[\frac{n_{2}}{y}-\frac{n_{1}}{x}-\frac{n_{2}-n_{1}}{r}\right] \theta^{2} \end{aligned} $$. Also show that the above expression leads to the paraxial image point which is consistent with Eq. (3.10); we may note that $u, v$ and $R$ are all negative quantities because they are on the left of the refracting surface.

Consider a point object $O$ in front of a concave refracting surface $S P M$ separating two media of refracting indices $n_{1}$ and $n_{2}$ (see Fig. 3.34); $C$ represents the center of curvature. In this case also one obtains a virtual image. Let $Q$ represent an arbitrary point on the axis. We now have to consider the optical path length $L_{\mathrm{op}}=n_{1} O S-n_{2} S Q$; show that it is given by
$$
\begin{aligned}
L_{\mathrm{op}} &=n_{1} O S-n_{2} S Q \\
& \approx n_{1} x-n_{2} y-\frac{1}{2} r^{2}\left[\frac{n_{2}}{y}-\frac{n_{1}}{x}-\frac{n_{2}-n_{1}}{r}\right] \theta^{2}
\end{aligned}
$$.
Also show that the above expression leads to the paraxial image point which is consistent with Eq. (3.10); we may note that $u, v$ and $R$ are all negative quantities because they are on the left of the refracting surface.

Key Concepts

Fermat's Principle

Fermat's principle states that the actual path taken by light between two points is the one that minimizes (or more generally makes stationary) the optical path length. This principle is used to derive the laws of reflection and refraction and provides the basis for analyzing the propagation of light through systems with varying refractive indices.

Concave Refracting Surface

A concave refracting surface is a curved interface (with an inward curvature) between two media of different refractive indices. The curvature, along with the change in refractive index, determines how light is refracted and whether the system produces real or virtual images, influencing the focusing and imaging properties of optical devices.

Optical Path Length

Optical path length is the product of the refractive index and the physical distance traversed by a light ray in that medium, representing the phase delay accumulated along its path. This concept is fundamental in determining interference and image formation since it relates geometric distances to the optical behavior of light in different media.

Paraxial Approximation

The paraxial approximation assumes that light rays make very small angles with the optical axis, allowing the simplification of trigonometric functions (for example, sin? ? ? and cos? ? 1 - ?²/2). This approximation simplifies the mathematics in optical systems, making it easier to derive imaging equations and analyze the behavior of light near the axis.

Virtual Image Formation

Virtual image formation occurs when light rays diverge after refraction or reflection, so that their backward extensions appear to converge at a point behind the optical element. Such images cannot be projected onto a screen, but they are perceived by an observer and are key in understanding image formation in many optical instruments.

Transcript

00:01 All right, so let's solve this optics problem, right? and we will reply that the optical path length, which is op, this which is n1 or times os minus into time skew.

00:23 And this n1 into are two index of repractions, right? these are indices of repraction, right? and we are required to show in verify that this optical path length is equal to this equation, right? so this is a mirror, right? this is a concave refracting surface, right? this is a concave refracting surface and this is point q, point c and this is theta, right? and this is the object or the point o right.

01:08 And this distance from c to p is r and this is from q to p is y and from o to p is x right.

01:19 And we must need to know that uvr are negative quantities right and we will also show that the optical path length leads to paroxial image point right so to reply all these we will apply some paroxial equation and we will use the permat sequence right so so we have found it is easy, we can easily pound that os is equal to x plus r square minus x r divided by x into theta square divided by two right and this is always distance and we have found sq distance right so is square distance could be pound and it is y plus r square minus yr divided by y times theta square divided by two right so we have found these two distances and we know that the optical path length which is l o which is here it is equal to n1 os distance minus into is skewn distance right so we will put these two quantities in the optical path length right so optical path length is equal to this is equal to n1 this is n1 into x plus r square minus x are divided by x right into theta square divided by two right minus into into y plus r square right r square minus y are divided by y are divided by y right into theta squared divided by 2.

03:53 So we have put these two quantities in this equation, right? so after simplification, we got that this is n1x plus n1, right, into r square minus x are divided by x into theta square divided by 2 right and minus into y minus into r square minus y are divided by y right right into r squared by y right into theta square divided by 2 right so this l op is equal to n1 x minus into y minus 1 half times r square right into in 2 divided by y minus in 1 divided by x minus minus minus x minus minus in 2 minus in 2 minus in 2 minus in 1 divided by r right and times theta square right so we have just quoted these two quantities in the optical path length equation right and we know that the past derivative of optical path length with respect to theta so the past derivative of this optical path length with respect to theta is equal to 0...

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