Question

A Galilean telescope adjusted for a relaxed eye is 36.1cm long. If the objective lens has a focal length of 39.5cm , what is the magnification? Follow the sign conventions.

Answer 1

The magnification of the Galilean telescope adjusted for a relaxed eye is approximately -1.09. Simplifying, we find that the magnification of the Galilean telescope is approximately -1.09 (rounded to two decimal places).

Step-by-step explanation:

To calculate the magnification of a Galilean telescope adjusted for a relaxed eye, we can use the formula:

Magnification (M) = - (fobjective / feyepiece)
Given that the length of the telescope (L) is 36.1 cm and the focal length of the objective lens (fobjective) is 39.5 cm, we can determine the focal length of the eyepiece (feyepiece) using the formula:

L = fobjective - feyepiece
Combining the two formulas, we can find the magnification:
Magnification (M) = - (fobjective / (L - fobjective))

Substituting the values, we get:
Magnification (M) = - (39.5 cm / (36.1 cm - 39.5 cm))

Simplifying, we find that the magnification of the Galilean telescope is approximately -1.09 (rounded to two decimal places).

Answer 2

Magnification, m = 11.61

Step-by-step explanation:

Given that,

Object distance, u = -36.1 cm

Focal length of the objective lens, f = +39.5 cm

Let v is the image distance. Using the formula of lens equation, we get :

`(1)/(f)=(1)/(v)-(1)/(u)`

`(1)/(v)=(1)/(f)+(1)/(u)`

`(1)/(v)=(1)/(39.5)+(1)/(-36.1)`

v = -419.39 cm

Magnification of the lens is given by :

`m=(v)/(u)`

`m=(-419.39)/(-36.1)`

m = 11.61

So, the magnification of the Galilean telescope is 11.61. Hence, this is the required solution.