The Galilean Telescope. Figure 34.64 is a diagram of a...

The Galilean Telescope. Figure 34.64 is a diagram of a Galilean telescope, or opera glass, with both the object and its final image at infinity. The image $I$ serves as a virtual object for the eyepiece. The final image is virtual and erect. (a) Prove that the angular magnification is $M=-f_{1} | f_{2} .$ ( b) A Galilean telescope is to be constructed with the same objective lens as in Exercise $34.57 .$ What focal length should the eyepiece have if this telescope is to have the same magnitude of angular magnification as the one in Exercise 34.57$?$ (c) Compare the lengths of the telescopes.

Key Concepts

Galilean Telescope

A Galilean telescope is an optical instrument that uses a combination of a converging (objective) lens and a diverging (eyepiece) lens. The configuration is arranged so that the object and final image are at infinity, and the image formed by the objective lens acts as a virtual object for the eyepiece. This design results in a final image that is virtual and erect, making it suitable for applications where an upright image is preferred.

Angular Magnification

Angular magnification is the ratio of the angle subtended by the image when viewed through the telescope to the angle subtended by the object when viewed with the naked eye. In telescopes, particularly in the Galilean configuration, the angular magnification can be derived from the focal lengths of the objective and eyepiece lenses. For the Galilean telescope, it is given by the ratio of the focal length of the objective to the absolute value of the focal length of the eyepiece, with attention paid to the sign conventions.

Virtual Object and Virtual Image

In optical systems, a virtual image is one that cannot be projected onto a screen, as the rays diverge and only appear to originate from a location behind the lens. In the Galilean telescope, the image formed by the objective lens acts as a virtual object for the eyepiece. This concept is crucial because it affects how the eyepiece processes the light rays to produce the final virtual, erect image that the observer views.

Focal Length and Sign Conventions

Focal length is a fundamental property of a lens that quantifies its converging or diverging power. In a Galilean telescope, the objective lens has a positive focal length while the eyepiece has a negative focal length due to its diverging nature. The sign convention is essential in deriving formulas, such as the expression for angular magnification, and ensuring that calculations account for the proper orientation and behavior of light through each lens.

Telescope Length Determination

The overall physical length of the telescope is determined by the separation between the objective and the eyepiece lenses. In a Galilean telescope, this separation is not simply the sum of the focal lengths due to the diverging nature of the eyepiece. Instead, it is given by the difference between the focal length of the objective and the absolute value of the focal length of the eyepiece. This difference is critical in practical telescope design and affects the instrument’s portability and usability.

Transcript

00:01 So here in this problem we are given this figure and there are three parts in this equation in the a part we have to prove that the angular magnification m is equals to minus f1 over f2 this thing we have to prove now we know that the object distance u is equals to y dash over f1 where f1 is the focal length of the objective length and y is the height of the real image and the image distance u dash is equals to negative y dash over f2 where f2 is a focal length of i piece and y dash is the height of virtual image now the relation for angular magnification of telescope m is basically it's the ratio of u dash over u so we can plug the village here u dash we have this one so negative y dash over f2 divide by this one is y dash over f1 so from here we'll get minus f1 over f2 hence we have proved that the angular magnification m is equals to minus f1 over f2 so this was the solution for the a part now next we'll be solving the b part of the problem now next i will be solving the b part where we have to calculate the focal length of the i piece that is f2 we have to calculate now f2 will be equals to negative f1 over m from the relation that we obtained earlier we can write this thing now if you refer the question number 34 34 point 5 7 where the value of m is negative 6 .33 and f1 we have 95 centimeter so we can plug the values here to calculate f2 after will be equals to this is negative 95 divide by m here is in negative 6 .33 so from here if two comes out to be as 15 centimeter so this is the solution for the b part now next we'll be solving the c part now in the c part we have to compare the length of the telescope so the length of the telescope l over b basically we have to calculate the ratio of l over l0 so l we have f1 plus f2 and l not we have f1 minus f2 we have calculated all these value f1 we have 95 centimeter f2 we have 15 centimeter divide by f1 we have 95 centimeter and this one is 15 centimeter so from here we'll get the ratio as 1 .38 so this is the solution for the c part so i hope you have understood the problem...

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