Using the matrix method or otherwise, find the focal lengths and the location of the principal p

Using the matrix method or otherwise, find the focal lengths...

Key Concepts

Lensmaker's Equation

The lensmaker's equation relates the focal length of a lens to its curvature radii, refractive index, and thickness. It is useful for obtaining the theoretical focal length of a single lens or individual spherical surface combination. While ideal for thin lenses, in systems with appreciable thickness or multiple components, modifications or alternative methods like matrix analysis may be needed to account for additional complexities.

Principal Planes

Principal planes are conceptual surfaces in an optical system from which distances are measured to define object and image positions. They represent the locations where the light appears to refract in a complex lens or system of lenses. Determining the position of the principal planes is key for correctly applying the lens formula and understanding the imaging properties of the system.

Ray Transfer Matrix Method

This method uses matrices (often called ABCD matrices) to model the propagation of paraxial rays through an optical system. It simplifies the analysis by representing each optical element (such as curved surfaces, gaps, etc.) as a matrix that transforms the ray’s height and angle. By multiplying these matrices for a sequential optical setup, one can derive overall system properties like effective focal length and the positions of the principal planes.

Focal Length

The focal length of an optical system is a measure of its converging or diverging power. It is defined as the distance from the principal plane to the point where rays initially parallel to the optical axis converge (or appear to diverge from). In complex systems, the effective focal length can be found using analysis methods such as the ray transfer matrix or the lensmaker's equation applied appropriately.

Transcript

00:01 Okay, so in this problem we will be using matrices and let in part a, the powers of the two spherical surfaces are given by, so this is p1 which is equal to n prime minus n over r and this is equal to 1 .5 minus 1 over minus 1 and this is equal to minus 0 .5 right? and p2 is equal to n prime minus n over r and this is equal to 1 minus 1 .5 over r and this is equal to r is infinity so this is so this is equal to 1 minus 1 .5 over infinity and this is equal to 0 right so suppose that a parallel incidence ray which is a 1 this is equal to 0 strikes the print of the surface of the system at a height of y1 right and by using matrix method we can find the ray angle which is alpha 2 right and the height y2 prime right so at the back surface of the system it is given by we can find these quantities by matrices at the back side right so it is given by a two prior two bar y2 prime and this matrix this is matrix and this is equal to r2 times t1 2 r1 times the matrix of this right and this is equal to 1 p2 0 1 right so this is the entries of this and at this matrix we have one zero minus d1 prime bar right and this is one and times this matrix right in this matrices 1 p1 time 0 1 and times this matrix and times this matrix this is a 1 bar y1 right and so finally after putting the values we have this a 2 bar y2 prime so this matrix and this is equal to so this is 1 .01 right and i'm just putting the values here right this is 1 .0 minus 0 .31 right and this is 1 minus 0 .31 right and this is 1 minus 0 .31 right and this is 1 minus 0 .3.

03:02 Pipe and 0 1 right and this is 0 y1 right and this is minus 0 .5 y1 and this is 1 .1 and this is 1 .1 5 y1 right so the focal length is given by this is poker length and this is given by y1 divided by a 2 bar and this is equal to y 1 over 0 .5 .y1 and this is equal to y1 and this is equal to 2 meters right so the vocal length is 2 meter right and the principal plane is located at a distance of d to the left side of the right and surface of the system right so which is given by this d and this d is given by y2 prime minus y1 and modulus and divided by a 2 bar modulus and this is equal to 1 .1 pipe y1 minus y1 modulus divided by 0 .5 y1 right and this is equal to 0 .3 meters right so d is equal to 0 .3 meters right and in part b let the powers of the poor spherical surfaces are given by so the power of poor spherical surfaces are done by n prime minus in over r this is equal to 1 .5 minus 1 over infinity is equal to 0 so p 1 is equal to 0 right p 1 is equal to 0 and p 2 is equal to n m minus n prime over r this is equal to 1 -1 1 minus 1 .5 or minus 0 0 .5 and this is equal to 1 so p 2 is equal to 1 right and p 3 is equal to n prime minus n over r this is equal to 1 .5 minus 1 over minus 1 this is equal to minus 0 .5 so p 3 is equal to minus 0 .5 right and p 4 is equal to n minus n prime over r this is equal to 1 minus 1 .5 over infinity so this is equal to 0 so p 4 is equal to 0 right and suppose that we have a parallel ray which is n said and incident ray and this is alpha 1 right this is zero strikes at the front surface of the system at a height of y 1 and by using the matrix method we can find ray angle alpha 4 right and height y4 prime at the back side right so let find these values by matrix method which is y pore prime and this is equal to r4 t3 poor right r 3 t 2 3 3 r 2 t 1 right so these are matrices these are just matrices right so putting these matrices so this is a poor bar y4 prime is equal to 1 p poor right right 0 1 right time this matrix which is 1 0 minus d3 prime right and bar and this is 1 right times this matrix this is 1 p3 1 right and times this matrix and times this matrix this is a 1 0 this is 1 d2 prime bar this is minus right and this is 1 1 1 1 times this matrix this is one zero minus d1 prime bar right and one and time this matrix this is one p1 0 1 right and time this matrix which is a 1 alpha 1 bar and y1 right so putting the values we can find these values here so this is equal to one zero one right one zero one so this is an identity matrix right and this is a one zero minus 0 .151 right in this matrix which is one minus 0 .5 .01 right times this matrix this is one minus 0 .21 minus 0 .21 right and one one zero one right and this is one zero minus 0 .151 right and this is one zero one right this matrix and times this matrix this is zero and this is y1 right so solving these matrices by some method we got that a poor bar y 4 prime is equal to this is equal to 0 .6 y1 and this is equal to 0 .71 right so a poor bar is this and y4 prime is this right so the focal length is given by f is equal to y1 over a poor alpha 4 modulus right and this is equal to y1 over 0 .6 y1 right and this is equal to 1 .67 meters right so vocal length is 1 .67 meters okay and the principal plane right the principal plane is located at a distance d to the left side of the right end surface of the system so which given by so so d is equal to y4 prime minus y1 modulus over d alpha poor right.

09:46 So this is alpha -4...

Please give Ace some feedback